Determination of sulfate with Potentiometric Method

Titrimetric determination of sulfate

Summary
The present bulletin describes three potentiometric and one photometric titration method for the determination of sulfate. Which indication method is the most suitable depends above all on the sample matrix and is illustrated with examples in thisbulletin.

Method 1: Precipitation as barium sulfate and back-titration of the Ba2+ excess with EGTA. The ion-selective calcium electrode is used as indicator electrode.

Method 2: As in method 1, but with the electrode combination tungsten/platinum.

Method 3: Precipitation titration in semi-aqueous solution with lead perchlorate using the ion-selective lead electrode as indicator electrode.

Method 4: Photometric titration with barium perchlorate, thorin indicator and the 662 Photometer or 525 nm Spectrode. Particularly suitable for micro determinations!

Instruments and accessories

• 702 SET/MET Titrino, 716 DMS Titrino, 736 GP Titrino, 751 GPD Titrino or 785 DMP Titrino or796 Titroprocessor with 700 Dosino or 685 Dosimat

• 2.728.0040 Magnetic Stirrer

• 6.3014.223 Exchange unit(s) The necessary electrodes and other accessories are listed under the different methods.

Sample preparation

A) If calcium and magnesium ions interfere (method 2)
Sample matrix: mostly water samples Percolate 50 ... 100 mL of the aqueous sample to be analyzed through a strongly acidic cation exchanger column (e.g. Dowex 50) at a rate of 3 ... 4 drops/s. Discard the first 5 mL. The sulfate is then determined in a portion of the sample solution thus treated.

B) If chloride ions interfere (method 3)
In the aqueous sample the chloride is titrated with silver nitrate using the 6.0430.100 Ag Titrode. Percolate the titrated sample through a strongly acidic cation exchanger column and make up to 50 mL by rinsing with dist. water (do not discard the initial drops). 10 mL of the sample solution thus treated is then used for the sulfate titration.

C) Organically bound sulfur
Organically bound sulfur is converted to sulfate or sulfuric acid by means of a suitable digestion procedure (e.g. Wurzschmitt digestion, Schöniger combustion or Wickbold combustion).

Examples:

Combustion according to Schöniger

10 ... 30 mg sample is weighed exactly onto a sulfate-free filter paper (shape according to DIN 51400) and distributed evenly. Fit the filter paper into the platinum gauze cage of the ignition device. Pour 20 mL w(H2O2) = 10% into a 500 mL digestion flask, then fill the remaining space in the flask with pure oxygen. Fit the ignition device into the mouth of the digestion flask and ignite the sample. When combustion is finished, keep the flask closed and shake until the smoke has been completely absorbed by the liquid. Rinse out the liquid into a glass beaker with dist. water, add 1 mL c(NaOH) = 0.1 mol/L and evaporate to dryness. Take care also to heat the walls of the glass beaker, since residual peroxide can interfere with the sulfate determination. Dissolve the residue in 10 mL dist. water, then titrate the sulfate.

Wurzschmitt digestion

To avoid accidents, please adhere strictly to the instructions for use for the digestion apparatus! We have treated an organic substance (M = 500 g/mol, 1 S atom) as follows: 250 g sample is digested in the Wurzschmitt bomb with 250 mg ethylene glycol and 12 g sodium peroxide. After cooling down, the residue is dissolved in approx. 100 mL dist. water, boiled and allowed to cool down again. Neutralize with conc. HNO3 and make up to 250 mL with dist. water in a volumetric flask at 20 C. 1 ... 10 mL of this sample solution is then used for the sulfate titration. Using a sample weight of 500 mg the limit of quantitation is 0.008% S.

Method 1

General

Chloride ions present in the sample do not interfere with this method if their concentration is not too high. Samples like brines or sea water must be diluted. Calcium can be determined simultaneously. If no calcium ions are present, they can be added to the sample prior to the titration. Magnesium is not determined, but also does not interfere with the determination.

Electrodes

• 6.0504.100 ion-selective calcium electrode (Ca ISE) with 6.2104.020 electrode cable

• 6.0726.107 double-junction Ag/AgCl reference electrode [filled with c(KCl) = 3 mol/L] with 6.2106.020 electrode cable

• 
  

Reagents

• Barium chloride solution, c(BaCl2) = 0.05 mol/L: Dissolve 12.34 g BaCl2 * 2 H2O (99%) in c(HCl) = 0.1 mol/L and make up to 1 L.

• Titrant: c(EGTA) = 0.05 mol/L: Make a suspension with 19.4 g ethylene glycol-bis-(2-aminoethyl)-tetraacetic acid (98%) and 200 mL dist. water. Under stirring add c(NaOH) = 10 mol/L until everything has dissolved. Allow to cool down and make up to 1 L with dist. water.

• Ca2+ standard solution, c(CaCl2) = 0.1 mol/L, e.g. Metrohm no. 6.2301.070

• Buffer solution pH = 10: Dissolve 9 g NH4Cl and 60 mL w(NH3) = 25% in dist. water and make up to 1 L.

Analysis

Acidify the sample solution, if required, to pH <4>2) = 0.05 mol/L. If necessary, add 0.5 mL c(CaCl2) = 0.1 mol/L and allow to react for 3 min under stirring. Afterwards add 5 mL buffer solution pH = 10 and allow to react for another 30 s, then titrate with c(EGTA) = 0.05 mol/L. Two equivalence points are obtained, the first of which corresponds to the Ca2+ content and the difference between the second and the first equivalence point to the Ba2+ excess. The titrant consumption for the added quantity of c(BaCl2) = 0.05 mol/L has first to be determined. This is done by means of a blank sample (without sulfate), which is prepared and titrated in exactly the same way as the actual sample. The resulting blank consumption is stored as common variable C30 in the titrator.

Calculation

1 mL c(EGTA) = 0.05 mol/L corresponds to 4.803 mg SO4^2-or 1.603 mg S

2.004 mg Ca2+

RS1 = mg/L Ca2+ = EP1 * C01 * C02 / C00

RS2 = EP2 – EP1; mL

RS3 = mg/L SO4^2– = (C30 – RS2) * C03 * C02 / C00

EP1 = titrant consumption to reach the first EP in mL

EP2 = titrant consumption to reach t EP in mL

C00 = sample volume in mL

C01 = 2.004

C02 = 1000 (conversion factor in mL/L)

C03 = 4.803

C30 = blank consumption in mL [use the same quantity of c(BaCl2) = 0.05 mol/L for the blank as for the sample!]

Remarks

Alkaline sample solutions have to be acidified to pH <4>2.

If the sample contains no magnesium the expensive EGTA can be replaced by the cheaper titrant EDTA.

Method 2

General

Chloride ions do not interfere with this method. Calcium, however, does interfere as it is also titrated. It therefore has to be determined separately and deducted accordingly in the sulfate titration or removed from the sample solution by means of cation exchange.

Electrodes

• 6.1248.050 W electrode rod with 6.1241.030 electrode shaft and 6.2114.000 electrode cable

• 6.1248.000 Pt electrode rod with 6.1241.030 electrode shaft and 6.2106.020 electrode cable or 6.0726.107 double-junction Ag/AgCl reference Electrode [filled with c(KCl) = 3 mol/L] with 6.2106.020 electrode cable

Reagents

As described under method 1, except for the Ca2+ standard solution.

Analysis

Acidify the sample solution to pH <4>2) = 0.05 mol/L. Allow to react for 3 min under stirring. Afterwardsadd 5 ... 10 mL buffer solution pH = 10 and titrate back the Ba2+ excess with c(EGTA) = 0.05 mol/L using the MET mode (volume increment: 0.1 mL, fixed waiting time: 20 s). The titrant consumption for the added quantity of c(BaCl2) = 0.05 mol/L has first to be determined. This is done by means of a blank sample (without sulfate), which is prepared and titrated in exactly the same way as the actual sample. The resulting blank consumption is stored as common variable C30 in the titrator.

Calculation

1 mL c(EGTA) = 0.05 mol/L corresponds to 4.803 mg SO4^2– or 1.603 mg S

mg/L SO4^2– = (C30 – EP1) * C01 * C02 / C00

EP1 = titrant consumption in mL

C00 = sample volume in mL

C01 = 4.803

C02 = 1000 (conversion factor in mL/L)

C30 = blank consumption in mL

Remarks

• The sample solution has to be acidified to pH <4>2.

• For the blank the same quantity of c(BaCl2) = 0.05 mol/L has to be used as for the sample.

• With the Pt electrode steeper and better titration curves are obtained than with the Ag/AgCl reference electrode.

Method 3

General

Chloride, hydrogen carbonate and carbonate ions interfere with the determination. The sample should contain no acetate ions as these can severely affect the response of the ion-selective lead electrode.

Electrodes

• 6.0502.170 ion-selective lead electrode (Pb ISE) with 6.2104.020 electrode cable

• 6.0808.000 glassy carbon rod electrode with 6.2106.020 electrode cable (as reference electrode)

Reagents

Titrant: lead perchlorate solution, c[Pb(ClO4)2] = 0.005 mol/L in (isopropanol) = 85% (volume fraction): Made from Pb2+ standard solution c[Pb(ClO4)2] = 0.1 mol/L, Metrohm no. 6.2301.050

Perchloric acid, c(HClO4) = 1 mol/L

Isopropanol, p.a.

Analysis

To 10.0 mL chloride-free sample solution add 70 mL isopropanol and 1 mL c(HClO4) = 1 mol/L and titrate with c[Pb(ClO4)2] = 0.005 mol/L using the MET mode (volume increment: 0.1 mL, fixed waiting time: 30 s).

Calculation

1 mL c[Pb(ClO4)2] = 0.005 mol/L corresponds to 0.4803 mg SO4^2– or 0.1603 mg S

mg/L SO4^2– = EP1 * C01 * C02 / C00

EP1 = titrant consumption in mL

C00 = 10.0 (sample volume in mL)

C01 = 0.4803

C02 = 1000 (conversion factor in mL/L)

Remark

The surface of the Pb ISE has to be polished from time to time with moist aluminum oxide powder (6.2802.000 polishing set).

Method 4

General

As this is a precipitation titration with photometric endpoint indication, only micro determinations should be carried out. (Greater sulfate contents will cause interferences due to precipitation, which also occurs on the mirror of the light guide.) Lead perchlorate should not be used as titrant because then chloride ions would interfere with the determination.

Accessories

662 Photometer including 6.1108.010 light guide or 6.5501.00X Spectrode 525 nm

Reagents

• Titrant: barium perchlorate solution, c[Ba(ClO4)2] = 0.005 mol/L in (isopropanol) = 85% (volume fraction)

• Indicator solution: 100 mg thorin in 100 mL dist. water

• Perchloric acid, c(HClO4) = 1 mol/L

• Isopropanol, p.a.

• 
  

Analysis

To 10.0 mL sample solution add 0.3 mL thorin indicator, 1 mL c(HClO4) = 1 mol/L and 70 mL isopropanol. Degas under vacuum for 30 s, then titrate with c[Ba(ClO4)2] = 0.005 mol/L using the MET mode (volume increment: 0.1 mL, fixed waiting time: 20 s). When working with the 662 Photometer, the transmission is adjusted to 80% at a wavelength of 520 nm prior to the titration.

Calculation

1 mL c[Ba(ClO4)2] = 0.005 mol/L corresponds to 0.4803 mg SO4^2– or 0.1603 mg S

mg/L SO4^2– = EP1 * C01 * C02 / C00

EP1 = titrant consumption in mL

C00 = 10.0 (sample volume in mL)

C01 = 0.4803

C02 = 1000 (conversion factor in mL/L)

Remark

The titer of the barium perchlorate solution diminishes with time and therefore has to be checked regularly [e.g. with c(H2SO4) = 0.005 mol/L].

Answer for Method 1

751 GPD Titrino 01106 751.0020

user th

card label:Appl.751

U(init) 25 mV MET U Sulfat W

smpl size 50 ml

EP1 2.150 ml -8 mV

EP2 7.317 ml -92 mV

SO4 2- 4.131 mg/L

stop V reached

============

Answer for Method 2

751 GPD Titrino 01106 751.0020

user th

card label:Appl.751

U(init) -464 mV MET U Sulfat Z

smpl size 0.04173 g

EP1 5.119 ml -492 mV

SO4 2- 10.474 g/kg

stop V reached

============

Answer for Method 3

751 GPD Titrino 01106 751.0020

user th

card label:Appl.751

U(init) -401 mV MET U Sulfat S

smpl size 3 ml

EP1 6.390 ml -350 mV

SO4 2- 10.650 mmol/l

SO4 2- 1023.0 mg/L

stop V reached

============

Answer for Method 4

751 GPD Titrino 01106 751.0020

user th

card label:Appl.751

U(init) 220 mV MET U Sulfat S

smpl size 3.0 ml

EP1 6.193 ml 197 mV

SO4 2- 10.322 mmol/l

SO4 2- 991.5 mg/L

stop V reached

============

Literature :

Metrohm, Application Bulletin No. 140/3 e

Chloride titrations with potentiometric end-point detection

Chloride titrations with potentiometric end-point detection

Summary
Together with acid/base titrations, the titrimetric chloride determination is one of the most frequently used titrimetric methods of analysis. It is employed more or less frequently in practically every laboratory. The purpose of this Bulletin is to demonstrate the possibilities for determining chloride in a wide range of concentrations using automatic titrators. Silver nitrate is normally used as titrant. (For ecological reasons one should refrain from using mercury nitrate). The titrant concentration depends on the chloride content of the sample to be analysed. It is especially important to choose the correct electrode for samples with low chloride contents.

Apparatus and accessories
• Titrino or Titrando with Dosino or Dosimate
• Magnetic Swing-out Stirrer
• Exchange unit
• 6.0726.100 Reference electrode (outer electrolyte KNO3, for use with separate Ag electrodes) with 6.2106.020 electrode cable
• Measuring electrodes; a wide range of suitable electrodes is available at Metrohm. Here a selection of these:
- 6.0430.100 Ag-Titrode **
-
- 6.0450.100 comb. Ag-ring electrode **
-
- 6.0331.010 Ag-rod electrode **
-
- 6.0350.100 Ag-ring electrode **
-
- 6.0502.120 AgCl-ISE
-
- 6.0502.180 Ag/S-ISE
-
** if desired bare or with AgCl resp. Ag2S coating.

Reagents
• Titrant: c(AgNO3) = 0.001 ... 0.1 mol/L
• Standard: c(KCl) = 0.1000 mol/L, e.g. Metrohm No. 6.2301.060, or dilutions from it Acid: c(HNO3) = 2 mol/L or c(H2SO4) = 1 mol/L

and for special applications:
• Acetone, p.a. as free from chloride as possible
• Acetic acid, w = 80 %, containing 1.9 g/L amidosulphamic acid
• Protective colloid: polyvinyl alcohol, e.g. Merck No. 114266 as 0.2 % aqueous solution (dissolve in hot dist. water)

General remarks
Silver nitrate with many anions causes more or less soluble precipitations. Thus, in mixtures of several anions, also several end-points can appear in the titration curve. Here, the anion causing the most insoluble precipitation is recorded first. Consequently, in a mixture of chloride, bromide and iodide, chloride would be titrated last. For the quantitative separation of mixtures, the solubility products of the Ag compounds must be as far apart as possible. In addition, no great differences in concentration should exist. In some cases, adding barium acetate and/or acetone can facilitate separation. Generally titration should take place in an acidic solution (acidify with HNO3 or H2SO4). Before the chloride determination, cyanide, sulphide and thiosulphate should be removed by means of oxidation, e.g. with H2O2. With samples containing peroxides (e.g. after digestions), these must first be destroyed before titration. For the determination of high chloride concentrations (in brines, salts), the sample is weighed, diluted with dist. water to a certain volume and a portion of this (aliquot) is titrated. To prevent an accumulation of the AgCl precipitation, protective colloid can be added to the sample solution. Polyvinyl alcohol (5 mL) 0.2 % per 100 mL sample solution prevents inclusions and keeps the electrode surface practically free from precipitation.
We prefer use of the Ag-Titrode. No electrolytes have to be replenished, nor will the diaphragms become blocked. For the titration of samples with small contents of chloride or of chlorides in aggressive solutions, we recommend use of an electrode coated with Ag2S.

General titration procedure
Place sample or an aliquot of this in a glass beaker and add 0.5 mL HNO3 or H2SO4. If content of chloride is high, dilute with dist. water 50 ... 100 mL. Immerse the electrode(s) and titrate on the mV measuring range with the selected AgNO3 concentration.

Calculation

1 mL c(AgNO3) = 0.1 mol/L = 3.5453 mg Cl- or 5.8443 mg NaCl or 7.4555 mg KCl

A few chosen examples

1. Chloride in drinking water
To 100 mL drinking water add 0.5 mL c(HNO3) = 2 mol/L and titrate with c(AgNO3)= 0.01 mol/L. Ag-Titrode with Ag2S coating.

mg/L chloride = EP1 x 0.3545 x 1000 / 100 = EP1 x 3.545

2. Chloride in dialysis and/or infusion solutions
To 5.0 mL sample add 2 mL c(HNO3) = 2 mol/L and 30 ... 50 mL dist. H2O. Titrate with c(AgNO3) = 0.1 mol/L using the Ag-Titrode.

mmol/L chloride = EP1 x 0.1 x 1000 / 5 = EP1 x 20

3. Chloride in Cr(VI)-bath
In a glass beaker pipette 5.0 mL bath sample and 20 mL each dist. H2O and ethanol. Add 0.5 mL conc. H2SO4, heat and boil for 5 min to convert Cr(VI) entirely to Cr(III). After cooling, titrate with c(AgNO3) = 0.01 mol/L. Use Ag-Titrode with Ag2S coating.

mg/L chloride = EP1 x 0.355 x 1000 / 5 = EP1 x 71

4. Chloride in acidic copper bath
In a glass beaker pipette 20.0 mL bath sample, 2 mL c(HNO3) = 2 mol/L and 50 mL dist. H2O. Using the Ag-Titrode, titrate with c(AgNO3) = 0.01 mol/L.

mg/L chloride = EP1 x 0.355 x 1000 / 20 = EP1 x 17.75

5. Chloride in nickel (sulphate/sulphamate) bath
Depending on the expected chloride content, pipette 1.0 ... 5.0 mL sample into a glass beaker. Add approx. 50 mL dist. H2O and 2 mL c(HNO3) = 2 mol/L and titrate using the Ag-Titrode with c(AgNO3) = 0.1 mol/L.

g/L chloride = EP1 x 3.5453 / C00

C00 = mL sample size

6. Chloride traces in cement and clinker
Weigh exactly 2.500 g sample into a glass beaker and mix to a slurry with 30 mL dist. H2O. Stirring, carefully add 6 mL conc. HNO3 and place beaker for 1 - 2 min in an ultrasonic bath. Filter through a paper filter (free from chloride) into a 100 mL graduated flask, rinse filter well with dist. H2O, fill up to mark and mix.Pipette 50.0 mL (corresponding to 1.25 g original sample) into a glass beaker, add 20 mL glacial acetic acid and approx. 0.5 g sodium acetate and titrate with c(AgNO3) = 0.01 mol/L using the MET-mode of the titrator.

% chloride = EP1 x 0.355 x 0.1 / 1.25 = EP1 x 0.0284
C00 = 1.25 g

7. Salt content in meats (Dried meat, sausage, ham, smoked fish etc.)
Cut the sample in tiny pieces with a knife. Weigh exactly 10 g of sample into a mixer together with 190 g dist. water and let run for 1 ... 2 min until the mixture is homogeneous.
Weigh 50 g of this homogeneous mixture into a glass beaker, then add 50 mL dist. H2O and 2 mL c(HNO3) = 2 mol/L. Titrate with c(AgNO3) = 0.1 mol/L using the Ag-Titrode.

% NaCl = EP1 x 5.844 x 0.1 / C00

8. Absorbable halogenated hydrocarbons (AOX)
The analysis of absorbable halogenated hydrocarbons traces represents a special case. After combustion, the gases are caught in 80 % acetic acid with 1.9 g/L sulphamic acid and titrated with c(AgNO3) = 2 mmol/L in 80 % acetic acid.

Electrodes:
6.0331.010 with Ag2S coating
6.0726.100 Ag/AgCl reference electrode. Outer electrolyte is
c(NaOOCCH3) = 2 mol/L in 80 % acetic acid

Literature
Metrohm Application Bulletin No. 130 / 2e

Potentiometric titration of calcium (magnesium) in dairy products

Potentiometric titration of calcium (magnesium) in dairy products

Summary
This bulletin describes a simple method for the determination of the calcium content in dairy products. Use of CuEGTA and the ion-selective copper electrode (Cu ISE) as indicator electrode allows the determination to be performed without time consuming sample preparation. If the complexing agent EDTA is used as titrant instead of EGTA, the sum of calcium and magnesium is obtained. The magnesium content can then be calculated from the difference between the results of the two titrations.



Theory
If the Cu ISE is used to indicate the complexometric titration of calcium, addition of copper ions is necessary. As the Cu2+ ions would also react with the titrant EGTA, they are already added to the sample in complexed form as CuEGTA. This complex reacts with the Ca2+ ions as follows: CuEGTA + Ca2+ CaEGTA + Cu2+

Instruments and accessories
· 702 SET/MET Titrino, 716 DMS Titrino, 736 GP Titrino, 751 GPD Titrino or 785 DMP Titrino or 726 or 796 Titroprocessor with 700 Dosino or 685 Dosimat
· 2.728.0040 Magnetic Stirrer
· 6.3014.XX3 Exchange Unit(s)
· 6.0502.140 ion-selective copper electrode (Cu ISE) with 6.2104.020 electrode cable
· 6.0726.100 double-junction Ag/AgCl reference electrode (bridge electrolyte KNO3 sat.) with 6.2106.020 electrode cable

Reagents
· Titrant: c(EGTA) = 0.1 mol/L: Dissolve 38.04 g ethylene glycol-bis-(2-aminoethyl)-tetraacetic acid in 250 mL c(NaOH) = 1 mol/L and, after cooling down, make up to 1 L with dist. water.
· Buffer solution pH = 10: Dissolve 54 g NH4Cl in approx. 400 mL dist. water, add 300 mL w(NH3) = 25% and make up to 1 L with dist. water.
· CuEGTA complex solution: Mix 100 mL c(EGTA) = 0.1 mol/L with 100 mL of a solution containing 0.2 mol/L NH4Cl and exactly 0.1 mol/L Cu(NO3)2. Titration can be used to check if this solution contains no excess Cu(II) or EGTA.
· Sulfuric acid, c(H2SO4) = 0.05 mol/L
· Sodium hydroxide solution, c(NaOH) = 0.1 mol/L

Sample preparation

Milk, milk drinks, yogurt, brine bath, etc.
Approx. 10 g sample is weighed exactly into a glass beaker and diluted with approx. 90 mL dist. water.
Cheese
Approx. 1 g finely grated cheese (calcium content approx. 1%) or an equivalent quantity of cheese with differing calcium content is weighed into a glass beaker. Add 10 mL c(H2SO4) = 0.05 mol/L and approx. 50 mL dist. water, then heat the
mixture to 40 °C and stir at this temperature for 10 min. After cooling down, neutralize to pH = 7 with c(NaOH) = 0.1 mol/L.

Analysis
Add 1 mL CuEGTA solution and 10 mL buffer solution pH = 10 to the prepared sample solution. Allow to react for 10 … 30 s under stirring, then titrate with c(EGTA) = 0.1 mol/L, using, e.g., the MET mode.

Calculation
1 mL c(EGTA) = 0.1 mol/L corresponds to 4.008 mg Ca2+

% Ca2+ = EP1 * C01 * C02 / C00

EP1 = titrant consumption in mL
C00 = sample weight in g
C01 = 4.008
C02 = 0.1 (conversion factor for %)

Answer

Ca2+ = 3.057 x 4.008 x 0.1 / 10.0091 = 0.122%

EP1 = 3.057 mL
C00 = 10.0091 g

Remarks
· If the magnesium content is also required, a second titration is carried out with c(EDTA) = 0.1 mol/L and addition of CuEDTA. This determines the sum of calcium and magnesium. The magnesium content can then be calculated from the difference between the titrant consumption of the two titrations: EP1(Mg) = EP1(Ca + Mg) – EP1(Ca) Use the same sample weight for both titrations! 1 mL c(EDTA) = 0.1 mol/L corresponds to 2.4305 mg Mg2+
· If for cheese, work is done without heating and without addition of H2SO4 (as for milk), it is not possible to determine the total calcium concentration (results found too low).
·
Literature
Metrohm Application Bulletin No. 235 / 2e

DETERMINATION OF CADMIUM WITH POTENTIOMETRIC METHOD

Titrimetric analysis of cadmium plating baths

Summary
This Bulletin describes titrimetric methods for the determination of cadmium, free sodium hydroxide, sodium carbonate and total cyanide. The free cyanide can be calculated from the total cyanide and the Cd content.

Apparatus and accessories
• Titrino or Titrando with Dosino or Dosimat
• Magnetic swing-out stirrer
• Exchange units
• Photometer (610 nm) or Cu ISE 6.0431.140 with Ag/AgCl reference electrode 6.0726.107 (KCl 3 mol/L) and electrode cable 6.2106.020
• Combined pH glass electrode 6.0255.100 with electrode cable 6.2104.020
• Ag Titrode with Ag2S coating 6.0430.100

Reagents
These are described under the individual analyses.

1. Determination of cadmium
1.1. By means of photometric titration (Photometer)
Reagents:
• c(Na2EDTA) = 0.1 mol/L
• w(formaldehyde) = 30%
• Buffer solution pH = 10:
Dissolve 114 mL w(NH3) = 25% and 14 g NH4Cl in dist. H2O and fill up to 200 mL.
• Colored indicator: Dissolve 100 mg each eriochrome black T and vitamin C in dist. H2O and fill up to 100 mL.
• w(KCN) = 6.5%

Analysis:
Place a 1.0 ... 2.0 mL bath sample, containing approx. 50 mg Cd in a beaker, add 1 mL KCN, approx. 80 mL dist. H2O, 20 mL buffer solution pH = 10 and 0.25 mL colored indicator. Slowly add 4 mL formaldehyde and, while stirring allow to react for 1 min (to set Cd free from the cyanide complex). Finally titrate with c(Na2EDTA) = 0.1 mol/L using the Photometer.

Calculations:
1 mL c(Na2EDTA) = 0.1 mol/L = 11.241 mg Cd

g/L Cd = EP1 * C01 / C00

C00 = Sample size in mL = 2 mL
C01 = 11.241

Answer
Cd = 7.308 x 11.241 / 2 = 41.07 g/L

EP1 = 7.308 ml




1.2. By means of potentiometric titration (Cu ISE)
Reagents:
• c(Na2EDTA) = 0.1 mol/L
• Cu(NH4)2EDTA, c = 0.1 mol/L (Merck No. 105217)
• Buffer solution pH = 10; see under 1.1
• w(HNO3) = 65%

Analysis:
Work under a fume cupboard, toxic HCN is set free!!!
Put 5.0 mL bath sample into a Kjeldahl flask and add approx. 10 mL dist. H2O. Tilting back and forth, carefully add HNO3 until the solution becomes definitely acidic.
Heat up the solution under a fume cupboard and cook until all cyanide is destroyed and entirely removed. After cooling, rinse the solution with dist. H2O into a 50 Ml graduated flask, fill up to the mark and mix. Pipet 10.0 ... 20.0 mL of the treated sample solution (corresponding to 1 ... 2 mLoriginal bath) into a beaker and complete to approx. 40 mL with dist. H2O. Add 5 mL buffer solution pH = 10 and 1 mL Cu(NH4)2EDTA and, while stirring allow to react for 1 min. Finally titrate with c(Na2EDTA) = 0.1 mol/L in the MET mode of the titrator, using the Cu ISE.

Calculations:
1 mL c(Na2EDTA) = 0.1 mol/L = 11.241 mg Cd

g/L Cd = EP1 * C01 / C00

C00 = Sample size in mL = 2 mL
C01 = 11.241

Answer

Cd = 7.438 x 11.241 / 2 = 41.81 g/L

EP1 = 7.438 mL


2. Determination of free NaOH and carbonate
Reagents:
• c(HCl) = 1 mol/L
• w(BaCl2) = 25%

Analysis:
Put approx. 50 mL dist. H2O and 2.0 mL bath sample into a glass beaker. After addition of 5 mL BaCl2 solution, titrate with c(HCl) = 1 mol/L, using the pH glass electrode, until shortly after the second endpoint.

Calculations:
1 mL c(HCl) = 1 mol/L = 40.0 mg NaOH or 106.0 mg Na2CO3

g/L NaOH = EP1 * C01 / C00
g/L Na2CO3 = (EP2 - EP1) * C02 / C00

C00 = Sample size in mL = 2 mL
C01 = 40
C02 = 106

Answer

NaOH = 1.963 x 40 / 2 = 39.26 g/L
Na2CO3 =(3.409-1.963) x 106 / 2 = 76.64 g/L

EP1 = 1.963 mL
EP2 = 3.409 mL

Remarks:
• The titration has to be stopped after reaching the second endpoint, otherwise toxic HCN may be set free. It is best to work under a fume cupboard!

3. Determination of total cyanide

Reagents:
• c(AgNO3) = 0.1 mol/L
• c(NaOH) = 2 mol/L
• w(potassium iodide) = 10%

Analysis:
Place approx. 50 mL dist. H2O and 2 mL NaOH into a glass beaker. Then add 1.0 mL bath sample and 2 mL KI solution, and titrate with c(AgNO3) = 0.1 mol/L using the Ag Titrode.

Calculations:
1 mL c(AgNO3) = 0.1 mol/L = 5.204 mg CN or 9.802 mg NaCN or 13.024 mg KCN

g/L „cyanide“ = EP1 * C01 / C00

C00 = Sample size in mL = 2 mL
C01 = 5.204 or 9.802 or 13.024

Answer

„cyanide“ =10.019 x 5.204 / 2 = 26 .07 g/L

EP1 = 10.019 mL


4. Calculation of free cyanide
To calculate the free cyanide, the content of cyanide combined with Cd must be subtracted from the total cyanide content. According to the formula K2Cd(CN)4 this corresponds to 0.926 g CN per g Cd. Free cyanide; g/L CN = (g/L total-CN) - (g/L Cd * 0.926)

Literature
Metrohm Application Bulletin No. 93/2 e

DETERMINATION OF CADMIUM WITH POTENTIOMETRIC METHOD

Titrimetric analysis of cadmium plating baths

Summary
This Bulletin describes titrimetric methods for the determination of cadmium, free sodium hydroxide, sodium carbonate and total cyanide. The free cyanide can be calculated from the total cyanide and the Cd content.

Apparatus and accessories
• Titrino or Titrando with Dosino or Dosimat
• Magnetic swing-out stirrer
• Exchange units
• Photometer (610 nm) or Cu ISE 6.0431.140 with Ag/AgCl reference electrode 6.0726.107 (KCl 3 mol/L) and electrode cable 6.2106.020
• Combined pH glass electrode 6.0255.100 with electrode cable 6.2104.020
• Ag Titrode with Ag2S coating 6.0430.100

Reagents
These are described under the individual analyses.

1. Determination of cadmium
1.1. By means of photometric titration (Photometer)
Reagents:
• c(Na2EDTA) = 0.1 mol/L
• w(formaldehyde) = 30%
• Buffer solution pH = 10:
Dissolve 114 mL w(NH3) = 25% and 14 g NH4Cl in dist. H2O and fill up to 200 mL.
• Colored indicator: Dissolve 100 mg each eriochrome black T and vitamin C in dist. H2O and fill up to 100 mL.
• w(KCN) = 6.5%

Analysis:
Place a 1.0 ... 2.0 mL bath sample, containing approx. 50 mg Cd in a beaker, add 1 mL KCN, approx. 80 mL dist. H2O, 20 mL buffer solution pH = 10 and 0.25 mL colored indicator. Slowly add 4 mL formaldehyde and, while stirring allow to react for 1 min (to set Cd free from the cyanide complex). Finally titrate with c(Na2EDTA) = 0.1 mol/L using the Photometer.

Calculations:
1 mL c(Na2EDTA) = 0.1 mol/L = 11.241 mg Cd

g/L Cd = EP1 * C01 / C00

C00 = Sample size in mL = 2 mL
C01 = 11.241

Answer
Cd = 7.308 x 11.241 / 2 = 41.07 g/L

EP1 = 7.308 ml




1.2. By means of potentiometric titration (Cu ISE)
Reagents:
• c(Na2EDTA) = 0.1 mol/L
• Cu(NH4)2EDTA, c = 0.1 mol/L (Merck No. 105217)
• Buffer solution pH = 10; see under 1.1
• w(HNO3) = 65%

Analysis:
Work under a fume cupboard, toxic HCN is set free!!!
Put 5.0 mL bath sample into a Kjeldahl flask and add approx. 10 mL dist. H2O. Tilting back and forth, carefully add HNO3 until the solution becomes definitely acidic.
Heat up the solution under a fume cupboard and cook until all cyanide is destroyed and entirely removed. After cooling, rinse the solution with dist. H2O into a 50 Ml graduated flask, fill up to the mark and mix. Pipet 10.0 ... 20.0 mL of the treated sample solution (corresponding to 1 ... 2 mLoriginal bath) into a beaker and complete to approx. 40 mL with dist. H2O. Add 5 mL buffer solution pH = 10 and 1 mL Cu(NH4)2EDTA and, while stirring allow to react for 1 min. Finally titrate with c(Na2EDTA) = 0.1 mol/L in the MET mode of the titrator, using the Cu ISE.

Calculations:
1 mL c(Na2EDTA) = 0.1 mol/L = 11.241 mg Cd

g/L Cd = EP1 * C01 / C00

C00 = Sample size in mL = 2 mL
C01 = 11.241

Answer

Cd = 7.438 x 11.241 / 2 = 41.81 g/L

EP1 = 7.438 mL


2. Determination of free NaOH and carbonate
Reagents:
• c(HCl) = 1 mol/L
• w(BaCl2) = 25%

Analysis:
Put approx. 50 mL dist. H2O and 2.0 mL bath sample into a glass beaker. After addition of 5 mL BaCl2 solution, titrate with c(HCl) = 1 mol/L, using the pH glass electrode, until shortly after the second endpoint.

Calculations:
1 mL c(HCl) = 1 mol/L = 40.0 mg NaOH or 106.0 mg Na2CO3

g/L NaOH = EP1 * C01 / C00
g/L Na2CO3 = (EP2 - EP1) * C02 / C00

C00 = Sample size in mL = 2 mL
C01 = 40
C02 = 106

Answer

NaOH = 1.963 x 40 / 2 = 39.26 g/L
Na2CO3 =(3.409-1.963) x 106 / 2 = 76.64 g/L

EP1 = 1.963 mL
EP2 = 3.409 mL

Remarks:
• The titration has to be stopped after reaching the second endpoint, otherwise toxic HCN may be set free. It is best to work under a fume cupboard!

3. Determination of total cyanide

Reagents:
• c(AgNO3) = 0.1 mol/L
• c(NaOH) = 2 mol/L
• w(potassium iodide) = 10%

Analysis:
Place approx. 50 mL dist. H2O and 2 mL NaOH into a glass beaker. Then add 1.0 mL bath sample and 2 mL KI solution, and titrate with c(AgNO3) = 0.1 mol/L using the Ag Titrode.

Calculations:
1 mL c(AgNO3) = 0.1 mol/L = 5.204 mg CN or 9.802 mg NaCN or 13.024 mg KCN

g/L „cyanide“ = EP1 * C01 / C00

C00 = Sample size in mL = 2 mL
C01 = 5.204 or 9.802 or 13.024

Answer

„cyanide“ =10.019 x 5.204 / 2 = 26 .07 g/L

EP1 = 10.019 mL


4. Calculation of free cyanide
To calculate the free cyanide, the content of cyanide combined with Cd must be subtracted from the total cyanide content. According to the formula K2Cd(CN)4 this corresponds to 0.926 g CN per g Cd. Free cyanide; g/L CN = (g/L total-CN) - (g/L Cd * 0.926)

Literature
Metrohm Application Bulletin No. 93/2 e

DETERMINATION OF Ph, SULPHUR AND ASCORBIC ACID IN WINE WITH SIMPLE ANALYSIS

Simple wine analysis

Summary
This bulletin describes simple methods for wine analysis, namely for pH value, total titratable acid, free sulphurous acid, total sulphurous acid as well as ascorbic acid (vitamin C) and other reductones.

Instruments and accessories
• 691 or 780 pH Meter
• Dosimat with Magnetic Stirrer and Exchange Units
• 6.0258.000 Unitrode (combined pH glass electrode with built-in Pt 1000 temperature sensor)
• 6.0309.100 double Pt sheet electrode with 6.2104.080 electrode cable
• Possibly 6.0259.100 Unitrode and 6.1110.100 Pt 1000 temperature sensor with 6.2104.080 electrode cable (alternatively to 6.0258.000 Unitrode)

Reagents
• Buffer solution pH = 4.00 (Metrohm no. 6.2307.100)
• Buffer solution pH = 7.00 (Metrohm no. 6.2307.110)
• Electrolyte solution c(KCl) = 3 mol/L (Metrohm no. 6.2308.020)
• Stock solution c(NaOH) = 1 mol/L: Dissolve 40.0 g NaOH in CO2-free dist. water and make up to 1 L.
• Titrant c(NaOH) = 0.1 mol/L: The above stock solution is diluted 1 : 10 with CO2-free dist. water.
• Sulphuric acid w(H2SO4) = 25%
• Titrant iodide/iodate solution:
Dissolve 0.5573 g potassium iodate KIO3 (dried at a temperature not exceeding 150 °C) in 700 mL dist. water, add 3.5 g potassium iodide KI and make up to 1 L with dist. water.
• Potassium iodide KI puriss. p.a.
• Glyoxal solution w(glyoxal) = 40%:
80 g glyoxal are mixed with 100 mL dist. water. Adjust the pH value of this mixture to 7.0 with c(NaOH) = 1 mol/L and make up to 200 mL with dist. water. The solution has to be stored in the refrigerator using an amber bottle.





Analysis
1. pH value
• Connect the combined pH glass electrode to the pH Meter. Then calibrate with buffer solutions pH = 7.00 and pH = 4.00.
• Rinse the electrode with dist. water and immerse it in the wine sample. When the drift value is reached, read off or print out the pH value. Indication of the value to two decimal places.
• Example: pH value = 3.60

2. Total titratable acid
• The combined pH glass electrode to be used has already been calibrated prior to the pH measurement. Carbon dioxide must first be removed from the sample.
• Pipette 10 mL wine sample and 10 mL dist. water into a 50 mL beaker. Add a drop of a suitable anti-foaming agent (such as octanol) and heat to boiling point Immediately allow to cool, then immerse the electrode and titrate to pH = 7 with c(NaOH) = 0.1 mol/L while stirring. Let the titrant consumption be A mL.
• Calculation: Total acid = A x 0.75 (calculated as tartaric acid)
• Example: Consumption = 2.80 mL c(NaOH) = 0.1 mol/L
Total acid = 2.80 x 0.75 = 2.10 g/L

3. Free sulphurous acid
• Connect the double Pt sheet electrode to the Ipol input of the pH Meter.
• Pipette 50 mL sample into a 100 mL beaker and mix with ca. 1 g KI and 5 mL w(H2SO4) = 25%. Now slowly titrate with iodide/iodate solution until the indicated value shows a big change, and remains in this state for at least 5 s. If not, add more titrant until the conditions described are reached.* Let the titrant consumption be B mL.
• Calculation: Free sulphur dioxide = B x 10 (calculated as SO2)
• Example: Consumption = 2.31 mL iodide/iodate solution
Free sulphurous acid = 2.31 x 10 = 23.10 mg/L
• Brand-new electrodes or electrodes that have been out of use for a long time, may respond sluggishly. In such a case, they will need to be regenerated as follows: Short across the two plugs of the double platinum electrode with a wire and connect to the negative pole (cathode) of a 6 V battery. Connect the positive pole (anode) of the battery to an iron nail. Then immerse the electrode and the nail in dilute sulphuric acid solution that still contains some sulphurous acid and leave for 2 ... 3 min. After removal, rinse the electrode thoroughly with dist. water.

4. Total sulphurous acid
• The combined sulphurous acid must first be released by adding NaOH.
• Pipette 25 mL c(NaOH) = 1 mol/L and 50 mL sample into a 100 mL beaker. Place on the magnetic stirrer and stir gently for 15 min. Then add 10 mL w(H2SO4) = 25% and ca. 1 g KI and titrate with iodide/iodate solution under the same conditions as described in the section «Free sulphurous acid». The calculation is the same as above.
• Example: Consumption = 7.50 mL iodide/iodate solution Total sulphurous acid = 7.50 x 10 = 75.0 mg/L

5. Ascorbic acid (vitamin C) and other reductones
• Free sulphurous acid interferes with the determination and must therefore be bound by adding glyoxal.
• Mix 50 mL sample with 2 mL glyoxal solution and allow to stand for 5 min. Then add 5 mL w(H2SO4) = 25% (no addition of KI) and titrate with iodide/iodate solution as described under «Free sulphurous acid». Let the titrant consumption be C mL.
• Remark:
If the determination of the free sulphurous acid yields a result exceeding the maximum permitted value, check for the presence of ascorbic acid (vitamin C) according to the method described last. The real content of free sulphurous acid is then calculated as follows:

mg/L SO2 = (B – C) x 10

• The content of vitamin C is calculated as follows:
mg/L vitamin C = C x 27.5

Example
Fig. 1: Determination of the free sulphurous acid in a wine sample by manual titration with iodide/iodate solution.

LiteratureMetrohm, Application Bulletin No.139/2

POTENTIOMETRIC ANALYSIS OF TIN (Sn) PLATING BATHS

Potentiometric analysis of tin (Sn) plating baths

Summary
Potentiometric titration methods for the analysis of acid and alkaline tin plating baths are presented. Following methods are described: tin(II) / tin(IV) / total tin, free fluoroboric acid or free sulfuric acid, chloride in acidic tin baths, free hydroxide and carbonate in alkaline tin baths.

Apparatus and accessories
• Titrino or Titrando with Dosino or Dosimat
• Magnetic swing-out stirrer
• Exchange unit(s)
• Pt Titrode 6.0431.100 with electrode cable 6.2104.020
• Comb. pH glass electrode 6.0255.100
• Ag Titrode with Ag2S coating 6.0430.100

Reagents
These are described under the individual analyses.

1. Iodometric tin determination
To increase the accuracy of this analysis, 10.0 mL bath sample is pipetted into a 100 mL graduated flask, filled up to the mark with distilled water and mixed well.

Reagents:
• c(Iodine solution) = 0.05 mol/L
• w(HCl) = 36%
• Iron powder p.a.

1.1. Tin(II)
Add 15 mL HCl and 50 mL dist. H2O to 10.0 mL diluted sample (corresponding to 1 mL original bath) in a beaker and titrate with c(iodine solution) = 0.05 mol/L against the Pt Titrode.








Application Bulletin No. 90/2 e
Potentiometric analysis of tin plating baths Page 2

1.2. Tin(IV) and total tin
In compliance with the tin content, pipet between 10.0 - 50.0 mL of the diluted sample (1 - 5 mL original bath) into a wide-necked Erlenmeyer flask and add 50 mL HCI. Stirring well, add ca. 1 g iron powder in small portions and when reaction subsides, warm up until the iron powder is entirely dissolved. Cool immediately and titrate with c(iodine solution) = 0.05 mol/L against the Pt Titrode.

Calculations:
1 mL c(Iodine solution) = 0.05 mol/L = 5.9345 mg Sn

Sn = EP1 x C01 / C00 ;2; g/L

C00 = sample size in mL original bath = 1 mL
C01 = 5.9345

Answer

Sn = 3.03 x 5.9345 / 1 = 17.98 g/L

Where
EP1 = 3.03 mL


Application Bulletin No. 90/2 e
Potentiometric analysis of tin plating baths Page 3

2. Free fluoroboric acid or free sulfuric acid
Reagents:
• c(NaOH) = 1 mol/L
• Sodium sulfate p.a.

Analysis:
Dilute 10.0 mL bath sample to approx. 50 mL with dist. H2O in a beaker. While stirring dissolve approx. 5 g Na2SO4 and titrate afterwards with c(NaOH) = 1 mol/L against the comb. pH glass electrode. The flat potential jump at pH = approx. 3.2 is evaluated.


Calculations:
1 mL c(NaOH) = 1 mol/L = 87.81 mg HBF4 or 49.037 mg H2SO4

HBF4 = EP1 x C01 / C00 ;2; g/L
H2SO4 = EP1 x C02 / C00 ;2; g/L

C00 = Sample size in mL original sample = 5 mL
C01 = 87.81
C02 = 49.037

Answer

HBF4 = 9.932 x 87.81 / 5 = 174.43 g/L
H2SO4 = 9.932 x 49.037 / 5 = 97.41 g/L

EP1 = 9.932 mL

3. Chloride determination in acidic tin baths

Reagents:
• c(AgNO3) = 0.1 mol/L
• w(HNO3) = 65%

Application Bulletin No. 90/2 e
Potentiometric analysis of tin plating baths Page 4

Analysis:
Pipet 1.0 mL bath solution into a beaker and dilute to approx. 50 mL with dist. H2O. Add 2 mL HNO3 and titrate with c(AgNO3) = 0.1 mol/L against the Ag Titrode (Ag2S-coating).

Calculations:
1 mL c(AgNO3) = 0.1 mol/L = 3.5453 mg Chloride

Chloride = EP1 x C01 / C00 ;2; g/L

C00 = sample size in mL original bath = 1 mL
C01 = 3.5453

Answer

Chloride = 3.473 x 3.5453 / 1 = 12.31 g/L

EP1 = 3.473 mL


4. Free hydroxide and carbonate in alkaline baths

Reagents:
• c(HCl) = 1 mol/L
• w(BaCl2) = 25%

Analysis:
Add 50 mL BaCl2 to 10.0 mL bath sample in a wide-necked Erlenmeyer flask and boil for a short time. Allow to cool and slowly titrate the still warm solution with c(HCl) = 1 mol/L against the comb. pH glass electrode.

Calculations:
Two endpoints are obtained. The consumption up to EP1 corresponds to NaOH, between EP1 and EP2 to tin and between EP2 and EP3 to carbonate.

1 mL c(HCl) = 1 mol/L = 40.0 mg NaOH or 106.0 mg Na2CO3

NaOH = EP1 x C01 / C00 ;2; g/L
Na2CO3 = (EP3 - EP2) x C02 / C00 ;2; g/L

C00 = Sample size in mL original sample =10 mL
C01 = 40
C02 =106

Answer

NaOH = 6.911 x 40 / 10 = 27.54 g/L
Na2CO3 = (14.329-11.657) x 106 / 10 = 28.32 g/L

EP1 = 6.911
EP2 = 11.657
EP3 = 14.329

Literature
Metrohm, Application Bulletin No.90/2e

DETERMINATION OF CALCIUM AND MAGNESIUM IN WATER SAMPLE AND BEVERAGES BY COMPLEXOMETRIC TITRATION WITH POTENTIOMETRIC METHOD

DETERMINATION OF CALCIUM AND MAGNESIUM IN WATER SAMPLE AND BEVERAGES BY COMPLEXOMETRIC TITRATION WITH POTENTIOMETRIC METHOD


Summary
This procedure describes a potentiometric titration method for the simultaneous complexometric determination of the calcium and magnesium ions in a sample using Na2EDTA as titran. The ion-selective calcium electrode is used as indicator electrode. To achieve a better differentiation between the Ca and Mg potential jumps an auxiliary complexing agent is added thet additionally masks Fe3+ and Al3+. Fixing the pH value at pH = 8.5 prevents interverences that might to be caused by the precipitation of CaCO3.
The method described is suitable for the automatic determination of the different water hardnesses (carbonat, calcium, magnesium, total and permanent hardness) the determination of calcium and magnesium in beverages (fruit and vegetable juices, wine) is also dealt with.

Instruments :

· Titrino or Titroprocessor
· 6.0504.100 ion-selective calcium electrode
· 6.0733.100Ag/AgCl reference electrode
· 6.0239.100 combined pH glass electrode

Reagents :

· Titran for the determination of the carbonate hardness : c(HCl) = 0.1 mol/L.
· Titran for the Ca and Mg determination : c(Na2EDTA) = 0.05 mol/L + c(KOH) = 0.1 mol/L
· Auxiliary complexing solution : c(acetylacetone) = 0.1 mol/L + c(TRIS) = 0.2 mol/L (TRIS=tris(hydroxymethyl)aminomethane)
· Hydrochloric acid w(HCl) = 20% and sodium hydroxide w(NaOH) = 20%

Analysis :

· Water sample : determination of the water hardnesses

100 mL sample is placed in the titration vessel and titrated in a first titration with c(HCl) = 0.1 mol/L (combined pH glass electrode).
Using auxiliary dosimat, 15 mL of the auxiliary complexing solution is added and, after a short waiting time, the calcium and magnesium ions are titrated in a second titration with Na2EDTA (Ca ISE). The first equivalence point corresponds to the Ca 2+ content, the difference between the second and the first equivalence point to the Mg 2+ content.

Calculations

1 st titration :
Acid neutralizing capacity = EP1*0.1*1000/C00 ;2;mmol/L
Carbonat hardness = EP1*0.1*0.5*1000/C00 ;2;mmol/L (temporary hardness)

2 nd titration :
Calcium hardness = EP1*0.05*1000/C00 ;2;mmol/L
Magnesium hardness = (EP2-EP1)*0.05*1000/C00;2;mmol/L
Total hardness = EP2*0.05*1000/C00 ;2;mmol/L
Permanent hardness = (total hardness-carbonate hardness) ;2;mmol/L

Where
1 mmol Ca 2+ = 40.08 mg Ca 2+
1 mmol Mg 2+ = 24.31 Mg 2+
.
Answer

1 st titration :
Acid neutralizing capacity = 4.482x0.1x1000/100 = 4.48 mmol/L
Carbonat hardness = 4.482x0.1x0.5x1000/100 = 2.24 mmol/L (temporary hardness)

2 nd titration :
Calcium hardness = 4.482x0.05x1000/100 =2.24 mmol/L
Magnesium hardness =(6.078-4.482)*0.05*1000/100 = 0.80 mmol/L
Total hardness = 6.078*0.05*1000/100 = 3.04 mmol/L
Permanent hardness = (3.04-2.24) = 0.80 mmol/L

Where
EP1 = 4.482 mL
EP2 = 6.078 mL

· Beverages (fruit and vegetable juices, wine) : Determination of calcium and magnesium

25 mL sample is pipetted into an evaporating dish and evaporated in a drying cabinet at 140 oC. The residue is then heated to redmess in a muffle furnace at 600 oC, until a white ash remains. After cooling down, 2 mL w(HCl) = 20% is added, the mixture heated to dissolve the ash and the resulting solution rinsed into a glass beaker with dist. Water. The pH value of this sample solution is then adjusted to 8.5 with w(NaOH) = 20%. 20 mL of the auxiliary complexing solution is added and titration is performed with c(Na2EDTA) = 0.05 using the Ca ISE.

Calculation

1 mL c(Na2EDTA) = 0.05 mol/L corresponds tomg Ca 2+ or 1.216 mg Mg 2+.

Ca 2+ = EP1*2.004*1000/C00;2;mg/L
Mg 2+ = (EP2-EP1)*1.216*1000/C00 ;2;mg/L

EP1 = titrant consumption to reach the first EP in mL = 2.156 mL
EP2 = titrant consumption to reach the second EP in mL = 5.078 mL
C00 = 25.0 mL (sample volume in mL)

Answer

Ca 2+ = 2.156 x 2.004 x 1000/25 = 172.82 mg/L
Mg 2+ = (5.078-2.156) x 1.216 x 1000 /25 = 142.13 mg/L

Literature
Metrohm, Application Bulletin No.125/2e