Question

$0.001$ mol of the strong electrolyte $M(OH{)}_2$ has been dissolved to make a $20$ mL of its saturated solution. Its pH will be: $[K_w=1\times {10}^{-14}]$

• A. $13$
• B. $3.3$
• C. $11$
• D. $9.8$

Answer 1

The correct option is A $13$

Moles of $M{\left(OH\right)}_2=0.001$ mol

Volume of solution (V)=$0.02$

Therefore,

$M{\left(OH\right)}_2\to 2O{H}^{-}\left(aq\right)+M^{2+}\left(aq\right)$

Hence,

$0.001molesM{\left(OH\right)}_{2}produces2\times 0.001molesO{H}^{-}$

$\left[O{H_{aq}}^{-}\right]=\frac{0.002}{0.02}=0.1mol{L}^{-}$

$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$

$1\times {10}^{-14}=\left[H^{+}\right]\left[1\times {10}^{-1}\right]$

$pH=-\log \left[H^{+}\right]$

$=13$