Question

What is the minimum concentration of $KOH$ that must be added to $0.015MCaC{l}_2$ to create a precipitation?
Note: $K_{sp}$ of $Ca(OH{)}_2=4.68\times {10}^{-6}$

• A. $0.090MKOH$
• B. $0.075MKOH$
• C. $5.58\times {10}^{-3}MKOH$
• D. $3.12\times {10}^{-5}MKOH$

Answer 1

The correct option is A $0.090MKOH$

$Ca{\left(OH\right)}_2\rightleftharpoons {Ca}^{+2}+2{OH}^{-}$

$K_{sp}=\left[{Ca}^{+2}\right]\cdot {\left[2{OH}^{-}\right]}^2=4.68\times {10}^{-6}\left(\text{Given}\right)$

$\Rightarrow \left[{Ca}^{+2}\right]=0.015$

$\left[{OH}^{-}\right]=?$

Therefore,

$4.68\times {10}^{-6}=0.015\times 4\times {\left[{OH}^{-}\right]}^2$

$\Rightarrow \left[{OH}^{-}\right]=\sqrt{78\times {10}^{-6}}=8.83\times {10}^{-3}\approx 0.09$

Hence the minimum concentration of $KOH$ that must be added to $0.015MCa{Cl}_2$ to create a precipitation is $0.09M$.