Calculating the pH of a solution of Ca(OH)2 A solution prepared by dissolving $2.8 g$ of lime, $\ce{CaO}$ in enough water to make $1.00 l$ of lime water ($\ce{Ca(OH)_{2(aq)}}$). If solubility of $\ce{Ca(OH)2}$ in water is $1.48 g$. The $p\ce H$ of the solution obtained will be:

[$\log 2 = 0.3$, Atomic masses are $\ce Ca$ = $40$ , $\ce O$ = $16$, $\ce H$ = $1$]

[Note : $\ce {Ca(OH)2}$ is considered to be a weak base because  of its low solubility but it is 100% dissociated]

(A) $12.3$

(B) $12.6$

(C) $1.3$

(D) $13$

My attempt:As $2.8g$ of $\ce {CaO}$ dissolves in $1l$ water the molarity of the solution is 0.05M. So,the concentration of OH- ions in the solution is $0.1M$.So $p\ce {OH}$ is $1$ and hence $p\ce H$ is $13$. But, the answer is given to be $12.6$.What am I doing wrong?
You did not account for the fact that $\ce{Ca(OH)2}$ is partly precipitated. Firstly, to find out the moles of $\ce{Ca(OH)2}$ precipitated:-
The number of moles in solution is $\frac{1.48}{74}M$ which is $0.02M$. The total moles precipitated is just the moles of $\ce{CaO}$ minus $0.02M$ since the initial number of moles of 
$\ce{Ca(OH)2}$ is equal to the moles of $\ce{CaO}$ taken and hence the amount precipitated turns out to be $0.03M$. Nevertheless, this is not required. With $0.02M$ $\ce{Ca(OH)2}$ in solution, the $\ce{OH-}$ concentration turns out to be $0.04M$ which, assuming no contribution of $\ce{OH-}$ from water and complete ionization, gives a $p\ce H$of $12.6$.