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
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
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