Question

Calculate the solubility at 25 Celsius of CuBr in pure water and in 0.071 M NaI. Round your answer to 2 significant digits.

Solubility in pure water:

Solubility in 0.071 M NaI:

CuBr: 6.27 x 10^-9

Answer 1

CuBr(s) ============ Cu+(aq) + Br-(aq)

a) In case of pure water

Ksp = [Cu+][Br-] = s^2 = 6.27 * 10^(-9)

s = sqrt(6.27 * 10^(-9)) = 7.9 * 10^(-5) M

Hence the solubility of CuBr in pure water will be 7.9 * 10^(-5) M

b) 0.071M NaI

Since there is no common ion between CuBr and NaI, hence the final answer will be same as part (a)

Ksp = [Cu+][Br-] = s^2 = 6.27 * 10^(-9)

s = sqrt(6.27 * 10^(-9)) = 7.9 * 10^(-5) M

Hence the solubility of CuBr in pure water will be 7.9 * 10^(-5) M