Question

What is the silver ion concentration in a solution prepared by mixing 369 mL of 0.387 M silver nitrate with 405 mL of 0.417 M sodium chromate? The K_sp of silver chromate is 1.2 × 10^-12

Answer 1

moles of AgNO3 = 0.387 x 369/1000= 0.1428

moles of Na2CrO4 = 0.417 x 405 / 1000 = 0.169

2 AgNO3 + Na2CrO4 ----------------> Ag2CrO4 + 2 NaNO3

2 1 1 2

0.1428 0.169

here limiting reagent is AgNO3 .

Na2CrO4 required = 0.1428 x 1 / 2 = 0.0714

so excess Na2CrO4 = 0.169 - 0.0714 = 0.0976

concentration of [CrO42-] = 0.0976 / (0.405 + 0.369) = 0.1261 M

[CrO42-] = 0.1261 M

Ag2CrO4 <---------------> 2 Ag+ + CrO42-

Ksp = [Ag+]^2[CrO42-]

1.2 × 10^-12 =  [Ag+]^2 x 0.1261

[Ag+] = 3.08 x 10^-6 M

silver ion concentration [Ag+] = 3.08 x 10^-6 M