Question

A) The solubility product of Pbcl​2 is found to be 1.4 x 10 -4 at a temperature equal to the boiling point of water. Calculate the solubility of Pbcl2 at that temperature. B)Calculate the pH of the solution of KOH, NaOH with 0.006M.

Answer 1

Solubility =(1.4*10)-4
                =14 - 4
                =10
We know that the boiling point of water is 100° [degree Celsius (C°)]
Also, KOH is a base whose P<SUP>H</SUP> lies between 7 to 14
and the same is with NaOH who also lies between 7 to 14
As NaOH and KOH are strong bases, their Ph is 10.90 to 11 

I cannot solve further...
Sorry, but I tried my level best..
Do not mind as I study in 10th and have passed only S.A.-1 Yet..