The solubility product constant of Calcium hydroxide is 6.5 * 10^-6 . If 0.10 mo of sodium hydroxide is added to 1 L of 0.0001 M Ca(OH)2, what is the final concentration of the calcium ion? SHOW your work.
Again, I have no idea where to start
2 answers
Qsp = (Ca^2+)(OH^-)^2 = (1E-4)(0.1)^2 = 1E-6 < Ksp = 6.5E-6 so no ppt of Ca(OH)2 occurs and (Ca^2+) must be 1E-4M
This is a common ion problem where 0.10M NaOH is added into a solution containing 0.0001M Ca(OH)2.
Ca(OH)2 <=> Ca^+2 + 2OH^-
Ci: --- 0M 0.10M
∆Ci: --- +x +2x
Ceq: --- x 0.10+2x~0.10M
Ksp = [Ca^+2][OH^-]^2
6.5x10^-6 = [Ca^+2](0.10)^2
Solve for [Ca^+2] ...
= (6.5x10^-6)/(0.10)^2
= 6.5x10-4M
Ca(OH)2 <=> Ca^+2 + 2OH^-
Ci: --- 0M 0.10M
∆Ci: --- +x +2x
Ceq: --- x 0.10+2x~0.10M
Ksp = [Ca^+2][OH^-]^2
6.5x10^-6 = [Ca^+2](0.10)^2
Solve for [Ca^+2] ...
= (6.5x10^-6)/(0.10)^2
= 6.5x10-4M
1 answer