(c) 1.0cm^3 0.25 moldm^-3 potassium chromate(VI), K2CrO4, is added as an indicator.At the end-point, all the silver chloride has been precipitated in the 52.5cm^3 solution. A slight excess of silver ions causes a red-brown precipitate of silver chromate(VI) to form.2Ag + (aq) + CrO4 2- (aq) - - > Ag2CrO4 (s)(i) The solubility product of silver chloride is:Ksp = 2.0 ×10^-10 mol^2 dm^-6Use this value to explain why [Ag+] = 1.41 × 10^-5 moldm^-3 at the end point. 1.0cm 0.25mol dm potassium chromate(VI), K,Cro,, is added as an indicator.-3At the end-point, all the silver chloride has been precipitated in the 52.5 cm solution. A slightexcess 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%3Dsp.Use this value to explain why [Ag*] = 1.41 x 10 mol dm 3 at the end-point.

A solution is said to be saturated solution, when the exact amount of solute has been dissolved in…

Answered: 1.0cm3 0.25mol dm3 potassium…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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(c) 1.0cm^3 0.25 moldm^-3 potassium chromate(VI), K2CrO4, is added as an indicator. At the end-point, all the silver chloride has been precipitated in the 52.5cm^3 solution. A slight excess of silver ions causes a red-brown precipitate of silver chromate(VI) to form. 2Ag + (aq) + CrO4 2- (aq) - - > Ag2CrO4 (s) (i) The solubility product of silver chloride is: Ksp = 2.0 ×10^-10 mol^2 dm^-6 Use this value to explain why [Ag+] = 1.41 × 10^-5 moldm^-3 at the end point.

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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