Question

The solubility of ${Sr\left(OH\right)}_2$ at $298K$ is $6.05{gL}^{-1}$. The pH of this solution is: [Molecular weight of ${Sr\left(OH\right)}_2=121g/mol]$

• A. $1.3$
• B. $2.3$
• C. $12.7$
• D. $11.7$

Answer 1

Answer: (C)

$12.7$

Solubility of $Sr{\left(OH\right)}_2=6.05g{L}^{-1}$

Molecular weight of $Sr{\left(OH\right)}_2=121.63g$

$\therefore$ concentration of $Sr{\left(OH\right)}_2=\frac{6.05}{121.63}=0.0497M$

${Sr{\left(OH\right)}_2}_{(aq.)}⟶{{Sr}^{+2}}_{(aq.)}+2{{OH}^{-}}_{(aq.)}$

$\left[{Sr}^{+2}\right]=0.0497M$

$\left[{OH}^{-}\right]=2\times 0.0497=0.0994M$

As we know that,

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

$\Rightarrow {10}^{-14}=\left[H^{+}\right]\times 0.0994$

$\Rightarrow \left[H^{+}\right]=\frac{{10}^{-14}}{0.0994}\approx {10}^{-13}$

As we know that,

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

$\therefore pH=-\log {10}^{-13}$

$\Rightarrow pH=13\approx 12.7$

Hence the pH of the solution is 12.7