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 .

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thomasdecabooter thomasdecabooter · Answer 1 · 2020-03-06T03:50:12+01:00

Answer:

[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

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