$PH=11/3\Rightarrow \left[ {{H}^{+}} \right]={{10}^{-11/3}}={{10}^{-12}}\times {{10}^{0/7}}=5\times {{10}^{-12}}\Rightarrow \left[ {{H}^{+}} \right]\times \left[ O{{H}^{-}} \right]={{10}^{-14}}\Rightarrow \left[ O{{H}^{-}} \right]=2\times {{10}^{-3}}$  $Ca{{\left( OH \right)}_{2}}\to C{{a}^{2+}}+2O{{H}^{-}}$  $\left[ O{{H}^{-}} \right]=2x=2\left( a.M \right)=2\left( 1\times M \right)=2\times {{10}^{-3}}\Rightarrow M=0/001\Rightarrow molCa{{\left( OH \right)}_{2}}=0/001\times 0/2=2\times {{10}^{-4}}$ $2\times {{10}^{-4}}=\frac{y\times {{10}^{-3}}}{74}\times \frac{74}{100}\Rightarrow y=20mg$