Ask a New Question
Menu

Calculate the molar solubility of Cr(OH)3 in a solution with a PH of 11.6 knowing that the Ksp(Cr(OH)3)=6.7×10^-31.

1 answer

pH+pOH=14

14-pH=pOH

pOH=14-11.6=2.4

pOH=-log[OH^-]

[OH^-]=10^-[2.4]

[OH^-]=3.98 x 10^-3 M= initial concentration

Cr(OH)3 -------> Cr^3+ + 3OH^-

Ksp=[x][3.98 x 10^-3 M]^3

Ksp/[3.98 x 10^-3 M]^3=x

x=6.7×10^-31/[3.98 x 10^-3 M]^3

Solve for x
Similar Questions
  1. The solubility of Mn(OH)2 is 3.04 x 10^-4 gram per 100 ml of solution .A) write the balanced chem equation for Mn(OH)2 in
    1. answers icon 0 answers
  2. The solubility of Mn(OH)2 is 3.04 x 10^-4 gram per 100 ml of solution .A) write the balanced chem equation for Mn(OH)2 in
    1. answers icon 1 answer
  3. The solubility of Mn(OH)2 is 3.04 x 10^-4 gram per 100 ml of solution .A) write the balanced chem equation for Mn(OH)2 in
    1. answers icon 1 answer
  4. (Ksp= 5.0*10^-13)1) Calculate the molar solubility of AgBr in 3.0×10^−2 M AgNO3 solution. 2) calculate the molar solubility
    1. answers icon 3 answers
more similar questions