I'm stuck on this question:( 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.

At the end point all silver chloride is precipate out only that amount is present in the solution…

1.0cm 0.25 mol dm3 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. EAg*(aq) + Cro,2-(aq) → Ag,Cro̟(s) ) The solubility product of silver chloride is: K = 2.0 x 10-10 mol? dm 6 %3D Use this value to explain why [Ag*] = 1.41 x 10mol dm 3 at the end-point.

1.0cm 0.25 mol dm3 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. EAg(aq) + Cro,2-(aq) → Ag,Cro̟(s) ) The solubility product of silver chloride is: K = 2.0 x 10-10 mol? dm 6 %3D Use this value to explain why [Ag] = 1.41 x 10mol dm 3 at the end-point.

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter9: Aqueous Solutions And Chemical Equilibria

Section: Chapter Questions

Problem 9.8QAP

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