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4' 1.0cm3 0.25 mol dm 3 potassium chromate(VI), K,Cro, is added as an indicator. At the end-point, all the silver chloride has been precipitated in the 52.5 cm solution. A slight excess of silver idis causes a red-brown precipitate of silver chromate(VI) to form. 2Ag*(aq) + Cro,2-(aq)→ Ag,Cro,(s) (i) The solubility product of silver chloride is: = 2.0 x 10 10 mol? dm-6 KSP Use this value to explain why [Ag*] = 1.41 x 10 mol dm-3 at the end-point.

4' 1.0cm3 0.25 mol dm 3 potassium chromate(VI), K,Cro, is added as an indicator. At the end-point, all the silver chloride has been precipitated in the 52.5 cm solution. A slight excess of silver idis causes a red-brown precipitate of silver chromate(VI) to form. 2Ag*(aq) + Cro,2-(aq)→ Ag,Cro,(s) (i) The solubility product of silver chloride is: = 2.0 x 10 10 mol? dm-6 KSP Use this value to explain why [Ag*] = 1.41 x 10 mol dm-3 at the end-point.

Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter14: Chemical Equilibrium
Section: Chapter Questions
Problem 14.70QE
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1.0cm 0.25mol dm potassium chromate(VI), K,Cro,, is added as an indicator. -3 At the end-point, all the silver chloride has been precipitated in the 52.5 cm solution. A slight excess of silver iois causes a red-brown precipitate of silver chromate(VI) to form. 2Ag*(aq) + Cro,²-(aq) → Ag,CrO,(s) (1) The solubility product of silver chloride is: K = 2.0 x 10-10 mol? dm-6 %3D sp. Use this value to explain why [Ag*] = 1.41 x 10 mol dm 3 at the end-point.
Transcribed Image Text:1.0cm 0.25mol dm potassium chromate(VI), K,Cro,, is added as an indicator. -3 At the end-point, all the silver chloride has been precipitated in the 52.5 cm solution. A slight excess of silver iois causes a red-brown precipitate of silver chromate(VI) to form. 2Ag*(aq) + Cro,²-(aq) → Ag,CrO,(s) (1) The solubility product of silver chloride is: K = 2.0 x 10-10 mol? dm-6 %3D sp. Use this value to explain why [Ag*] = 1.41 x 10 mol dm 3 at the end-point.
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