# (1). Find the concentration of the weak acid HA given that K_a=10^(-5) and pOH = 11 (2). What is the concentration of OH^- in this solution?

Aug 31, 2016

$\textsf{\left(1\right) .}$

$\textsf{0.1 \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{\left(2\right) .}$

$\textsf{{10}^{- 11} \textcolor{w h i t e}{x} \text{mol/l}}$

#### Explanation:

$\textsf{\left(1\right)} .$

The auto - ionisation of water gives us:

$\textsf{p H + p O H = 14}$ at $\textsf{{25}^{\circ} C}$

From this we get:

$\textsf{p H = 14 - p O H = 14 - 11 = 3}$

This means that $\textsf{\left[{H}^{+}\right] = {10}^{- p H} = {10}^{- 3} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{H A}$ dissociates:

$\textsf{H A r i g h t \le f t h a r p \infty n s {H}^{+} + {A}^{-}}$

For which:

$\textsf{{K}_{a} = \frac{\left[{H}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]} = {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

The concentrations refer to those at equilibrium.

To find $\textsf{\left[H A\right]}$ I will set up an ICE table based on concentrations. Let $\textsf{C}$ be the initial concentration of $\textsf{H A} :$

$\textsf{\textcolor{w h i t e}{\times \times} H A \textcolor{w h i t e}{\times \times} r i g h t \le f t h a r p \infty n s \textcolor{w h i t e}{\times \times} {H}^{+} \textcolor{w h i t e}{\times x} + \textcolor{w h i t e}{\times \times} {A}^{-}}$

$\textsf{\textcolor{red}{I} \textcolor{w h i t e}{\times \times} C \textcolor{w h i t e}{\times \times \times \times \times \times} 0 \textcolor{w h i t e}{\times \times \times \times \times} 0}$

$\textsf{\textcolor{red}{C} \textcolor{w h i t e}{\times} - x \textcolor{w h i t e}{\times \times \times \times \times} + x \textcolor{w h i t e}{\times \times \times \times} + x}$

$\textsf{\textcolor{red}{E} \textcolor{w h i t e}{\times} \left(C - x\right) \textcolor{w h i t e}{\times \times \times \times \times} x \textcolor{w h i t e}{\times \times \times \times \times} x}$

From the expression for $\textsf{{K}_{a}}$ I can write:

$\textsf{{K}_{a} = \frac{{x}^{2}}{\left(C - x\right)} = {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

I have already found the value of $\textsf{x}$ which = $\textsf{\left[{H}^{+}\right] = 0.001 \textcolor{w h i t e}{x} \text{mol/l}}$.

Putting in that value$\textsf{\Rightarrow}$

$\textsf{\frac{{0.001}^{2}}{\left(C - 0.001\right)} = {10}^{- 5}}$

With a small value of $\textsf{{K}_{a}}$ like this it is common to assume that $\textsf{x}$ is much smaller than $\textsf{C}$ such that $\textsf{C - x}$ approximates to $\textsf{C}$.

However, in this case I don't know $\textsf{C}$ so I won't make that assumption.

So:

sf((10^(-6))/((C-10^(-3))$\textsf{= {10}^{-} 5}$

$\therefore$$\textsf{{10}^{- 5} \left(C - {10}^{- 3}\right) = {10}^{- 6}}$

$\therefore$$\textsf{\left(C - {10}^{- 3}\right) = {10}^{- 6} / {10}^{- 5} = 0.1}$

$\therefore$$\textsf{C = 0.1 + 0.001 = 0.1001 \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{C = \left[H A\right] = 0.1 \textcolor{w h i t e}{x} \text{mol/l"" "(1"sig.fig}}$

$\textsf{\left(2\right)} .$

Since we are told that:

$\textsf{p O H = 11}$

This means that:

$\textsf{- \log \left[O {H}^{-}\right] = 11}$

This gives:

$\textsf{\left[O {H}^{-}\right] = {10}^{- 11} \textcolor{w h i t e}{x} \text{mol/l}}$