Question

$50ml$ $H_{2}0$ is added to a $50ml$ solution of $Ba(OH{)}_2$ of strength $0.01M$. The pH value of resulting solution will be :

• A. 12.52
• B. 12.31
• C. 1.69
• D. none of these

Answer 1

Answer: (B)

12.31

$\because$ Solution is dilute,

$M_1{V}_1=M_2{V}_2$

$0.01\times 50=M_{2}100$

$M_2=\frac{0.01\times 50}{100}$

$=0.005M$

$Ba(OH{)}_2⟶B{a}^{2+}+2O{H}^{-}$

$1$ mol of $Ba(OH{)}_2⟶$$2$ moles of $O{H}^{-}$

$\therefore$ concentration of $O{H}^{-}$ would also be $2$ times of concentration of dilute $so{l}^n$ of $Ba(OH{)}_2$

$\left[O{H}^{-}\right]=2\times \left(0.005\right)$

$=0.01M$

$pOH=-\log \left[0.01\right]=-\log \left[{10}^{-2}\right]$

$pH=14-2=12$

So, value nearest to $12$ is $12.31$