Question

Equilibrium constant for the following equilibrium is given at $0^{o}C$
${Na}_2{HPO}_4\cdot {12H}_{2}O\left(s\right)\rightleftharpoons {Na}_2{HPO}_4\cdot {7H}_{2}O\left(s\right)+{5H}_{2}O\left(s\right)$

$K_p=31.25\times {10}^{-13}$
The partial pressure of water is:

• A. $\frac{1}{5}\times {10}^{-3}atm$
• B. $0.5\times {10}^{-3}atm$
• C. $5\times {10}^{-2}atm$
• D. $5\times {10}^{-3}atm$

Answer 1

The correct option is D $5\times {10}^{-3}atm$

Solution: $\left(D\right)$ $5\times {10}^{-3}atm$

${{Na}_{2}HP{O}_4.12H_{2}O}_{\left(s\right)}\rightleftharpoons {{Na}_{2}HP{O}_4.7H_{2}O}_{\left(s\right)}+5{H_{2}O}_{\left(g\right)};K_p=31.25\times {10}^{-13}$

$K_p={\left(P_{H_{2}O}\right)}^5$

$\Rightarrow {\left(P_{H_{2}O}\right)}^5=31.25\times {10}^{-13}$

$\Rightarrow P_{H_{2}O}=\sqrt[5]{53125\times {10}^{-15}}$

$\Rightarrow P_{H_{2}O}=5\times {10}^{-3}atm$

Hence the partial pressure of water is $5\times {10}^{-3}atm$.