Question

What is the concentration of $C{O}_3^{2-}$ in a $0.1MN{a}_{2}C{O}_3$ solution after the hydrolysis reactions have come to an equilibrium? ($K_a\left(HC{O}_3^{-}\right)=4.7\times {10}^{-11}$)

• A. $2\times {10}^4$
• B. $6.3\times {10}^{-4}$
• C. $2\times {10}^{-4}$
• D. $9.54\times {10}^{-2}$

Answer 1

The correct option is D $9.54\times {10}^{-2}$

The first step of hydrolysis of $C{O}_3^{2-}$ ions is predominant.
$\underset{0.1(1-h)}{C{O}_3^{2-}\left(aq\right)}+H_{2}O\left(l\right)\to \underset{0.1h}{HC{O}_3^{-}\left(aq\right)}+\underset{0.1h}{O{H}^{-}\left(aq\right)}$
$K_h=\frac{0.1h^2}{(1-h)}$

$K_h=\frac{K_w}{K_a}=\frac{1\times {10}^{-14}}{4.7\times {10}^{-11}}{K}_h=2.12\times {10}^{-4}$

Finding $\frac{C}{K_h}$

$\frac{C}{K_h}=\frac{0.1}{2.12\times {10}^{-4}}=471.7$
Since $\frac{C}{K_h}>100$ , we can use $1-h\approx 1$
$K_h=0.1\times h^{2}h=\sqrt{\frac{2.12\times {10}^{-4}}{0.1}}=0.046[C{O}_3^{2-}{]}_{eqb}=0.1\times (1-h)=0.1\times (1-0.046\left)\right[C{O}_3^{2-}{]}_{eqb}=9.54\times {10}^{-2}$