Asked by Last
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.
Answers
Answered by
Anonymous
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
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