I need help with this problem!!

Question

When after combining 74.00mL of 0.3948M strontium nitrate solution with 100.0mL of 0.4107M potassium chromate solution, yellow precipitate forms, SrCrO4(s). If Ksp(SrCrO4) = 3.571x10^-5, what are the concentrations of strontium ions and chromate ions remaining in the solution when the system reaches equilibrium? And how many grams of SrCrO4(s) would expect to form?

Answer 1

The critical reaction equation is CrO₄^2- + Sr²⁺ ↔ SrCrO₄ K = 1/K_SP we'll use the regular solubility reaction for the analysis with K = K_SP = [CrO₄^2-][Sr²⁺] at equilibrium. Molarity or moles can be used for the ICE table. Let's use moles because we ultimately want to find the grams of strontium chromate formed. So, let x be the moles of strontium chromate formed and the total volume of solution is .174 liters:

ICE diagram leads to the following changes x moles of SrCrO₄ formed, n_0,CrO4 - x moles of CrO₄^2- and n_0,Sr - x moles of Sr²⁺ at equilibrium.

note n_0,CrO4 = (.3948 mole/liter)(.074 liters) and n_Sr = (.4107 mole/liter)(.1 liters)

Plug all of that into the K expression and solve for x (Then multiply by the molar mass of SrCrO₄ to get grams.

K_SP = 3.571 x 10^-5 = (n_0,CrO4 - x)(n_0,Sr -x)/(.174 liters)² and solve quadratic for x

The concentrations of the reactants at equilibrium will be the (n_0,i -x)/.174