What is the silver ion concentration in a solution prepared by mixing 369 mL 0.373 M silver nitrate with 411 mL 0.401 M sodium chromate? The K sp of silver chromate is 1.2 * 10 - 12.

[A g + ] = 3.12 * 10^-6 M

Explanation:

Step 1: Data given

Volume silver nitrate = 369 mL

Molarity silver nitrate = 0.373 M

Volume sodium chromate = 411 mL

Molarity sodium chromate = 0.401 M

The Ksp of silver chromate is 1.2 * 10^ - 12

Step 2: The balanced equation

2 A g + (a q) + C rO4^2 - (a q) →Ag2CrO4 (s)

Step 3:

[Ag+]i = [AgNO3] * V1 / (V1+V2) * 1 mol Ag + / 1 mol AgNO3

[Ag+]i = 0.373 M * 0.369 / (0.369+0.411)

[Ag+]i = 0.176 M

[C rO4^2]i = [Ag2CrO4] * V2 / (V1+V2) * 1mol C rO4^2 / 1 mol Ag2CrO4

[C rO4^2i = 0.401 M * 0.411 / (0.369+0.411)

[C rO4^2]i = 0.211 M

Ksp = [Ag+]²[CrO4^2-] = 1.2 * 10^ - 12

[CrO4^2-]f = [CrO4^2-]i - 0.5 * [Ag+]i

[CrO4^2-]f = 0.211 - 0.088 = 0.123 M

1.2 * 10^ - 12 = (2 x) ² * (0.123M + x)

[ C O 3^ - 2] f >> x

1.2 * 10^ - 12 = (2 x) ² * 0.123M

x = 1.56 * 10^-6

[A g + ] f = 2x = 3.12 * 10^-6 M

,