# Question #873de

Jul 31, 2016

$\textsf{9.65}$

#### Explanation:

The $\textsf{H C {O}_{3}^{-}}$ ion is basic:

$\textsf{H C {O}_{3 \left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{2} C {O}_{3 \left(a q\right)} + O {H}_{\left(a q\right)}^{-}}$

$\textsf{{K}_{b} = \frac{\left[{H}_{2} C {O}_{3}\right] \left[O {H}^{-}\right]}{\left[H C {O}_{3}^{-}\right]} = 1.99 \times {10}^{- 8} \textcolor{w h i t e}{x} \text{mol/l}}$

We know that $\textsf{\left[{H}_{2} C {O}_{3}\right] = \left[O {H}^{-}\right]}$

Because the dissociation is small I will assume that the equilibrium concentration of $\textsf{\left[H C {O}_{3}^{-}\right]}$ is equal to $\textsf{0.1 \textcolor{w h i t e}{x} \text{mol/l}}$

So now we can write:

$\textsf{{K}_{b} = {\left[O {H}^{-}\right]}^{2} / \left(0.1\right) = 1.99 \times {10}^{- 8}}$

$\therefore$$\textsf{{\left[O {H}^{-}\right]}^{2} = 1.99 \times {10}^{- 8} \times 0.1 = 1.99 \times {10}^{- 9}}$

From which:

$\textsf{\left[O {H}^{-}\right] = 4.46 x {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

$\therefore$$\textsf{p O H = - \log \left(4.46 \times {10}^{- 5}\right) = 4.35}$

$\textsf{p H + p O H = 14}$

$\therefore$$\textsf{p H = 14 - 4.35 = 9.65}$