Titration
- Problem Set In Precipitation Acid Base And Complexometric Titration 1 (182.77 KiB) Viewed 119 times
Problem Set in Precipitation, Acid-Base and Complexometric Titration
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Problem Set in Precipitation, Acid-Base and Complexometric Titration
Problem Set in Precipitation, Acid-Base and Complexometric
Titration
4. Consider a titration for 50.0 mL of 0.0129 M Fe2+ with 0.015 M EDTA in a solution buffered to pH = 7.0. Calculate the pFe2+after the addition of 20.0 and 44.10 mL of the titrant and at the equivalence point. Kf = 2.1x1014 04- = 4.8x10-4 at pH 7.0 5. Calculate pCu2+ at each of the following points in titration of 50.0 mL of 0.010 M Cu2+ with 0.0125 M EDTA at pH 11 in a solution whose NH3 concentration is somehow fixed at 0.02 M. V = 35.0 ml, 40.0 mL and 45.0 mL K1 = 104.04 K2 = 103.43 K3 = 102.8 K4 = 101.48 04- = 0.81 at pH 11.0; Kf = 6.3x1018 6. (Indirect titration) Cyanide can be determined indirectly by EDTA titration. A known excess Ni2+ is added to cyanide to form tetracyanonickelate: 4CN+ Ni2+ + Ni(CN) When the excess Ni2+is titrated with standard EDTA, Ni(CN) - does not react. In a cyanide analysis, 12.7 ml of cyanide solution was treated with 25.0 mL of standard solution containing excess Ni2+ to form tetracyanonickelate. The excess Ni2+ required 10.1 mL of 0.0130 M EDTA for complete reaction. In a separate experiment, 39.3 mL of 0.0130 M EDTA was required to react with 30.0 mL of the standard Ni2+ solution. Calculate the molarity of CN in the 12.7-ml sample of unknown.
Titration
4. Consider a titration for 50.0 mL of 0.0129 M Fe2+ with 0.015 M EDTA in a solution buffered to pH = 7.0. Calculate the pFe2+after the addition of 20.0 and 44.10 mL of the titrant and at the equivalence point. Kf = 2.1x1014 04- = 4.8x10-4 at pH 7.0 5. Calculate pCu2+ at each of the following points in titration of 50.0 mL of 0.010 M Cu2+ with 0.0125 M EDTA at pH 11 in a solution whose NH3 concentration is somehow fixed at 0.02 M. V = 35.0 ml, 40.0 mL and 45.0 mL K1 = 104.04 K2 = 103.43 K3 = 102.8 K4 = 101.48 04- = 0.81 at pH 11.0; Kf = 6.3x1018 6. (Indirect titration) Cyanide can be determined indirectly by EDTA titration. A known excess Ni2+ is added to cyanide to form tetracyanonickelate: 4CN+ Ni2+ + Ni(CN) When the excess Ni2+is titrated with standard EDTA, Ni(CN) - does not react. In a cyanide analysis, 12.7 ml of cyanide solution was treated with 25.0 mL of standard solution containing excess Ni2+ to form tetracyanonickelate. The excess Ni2+ required 10.1 mL of 0.0130 M EDTA for complete reaction. In a separate experiment, 39.3 mL of 0.0130 M EDTA was required to react with 30.0 mL of the standard Ni2+ solution. Calculate the molarity of CN in the 12.7-ml sample of unknown.