M(OH)2---> M^+ + 2OH
14-pH=pOH
14-10.49=3.51
pOH=-log[OH]
10^(-3.51)=OH concentration
OH concentration is twice as much as M from the reaction.
Ksp=products/reactants=[OH Concentration]^2[OH concentration/2]
I calculate 1.48 x 10^-11
At 22 °C an excess amount of a generic metal hydroxide M(OH)2, is mixed with pure water. The resulting equilibrium solution has a pH of 10.49. What is the Ksp of the salt at 22 °C?
Is the ksp=1.38*10^-11?
2 answers
I agree with 1.48E-11
1 answer