# Aweak acid HA, has a pKa of 5.0. If 1.0 mol of this acid and 0.1 mol of NaOH were dissolved in one liter of water, what would the final pH be?

Sep 5, 2016

$\textsf{p H = 4.04}$

#### Explanation:

The weak acid is partly neutralised by the alkali:

$\textsf{H {A}_{\left(a q\right)} + N a O {H}_{\left(a q\right)} \rightarrow N a {A}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)}}$

Since you can see that they react in a 1:1 molar ratio we can say that the number of moles of HA remaining is 1 - 0.1 = 0.9.

This means that the number of moles of NaA formed is 0.1.

Effectively we have created an acidic buffer which consists of a mixture of a weak acid and its co - base.

We can find the pH using the expression for $\textsf{{K}_{a}}$.

HA is a weak acid and dissociates:

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

For which:

$\textsf{{K}_{a} = \frac{\left[{H}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]} = {10}^{- p {K}_{a}} = {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l"" } \textcolor{red}{\left(1\right)}}$

$\therefore$$\textsf{\left[{H}^{+}\right] = {K}_{a} \times \frac{\left[H A\right]}{\left[{A}^{-}\right]} \text{ } \textcolor{red}{\left(2\right)}}$

Note that these are equilibrium concentrations.

To find the concentrations at equilibrium we can set up an ICE table:

$\textsf{\text{ "HA" "rightleftharpoons" "H^+" " +" } {A}^{-}}$

$\textsf{\textcolor{red}{I} \text{ "0.9" "0" } 0.1}$

$\textsf{\textcolor{red}{C} \text{ "-x" "+x" } + x}$

$\textsf{\textcolor{red}{E} \text{ "(0.9-x)" "x" } \left(0.1 + x\right)}$

$\therefore$$\textsf{\frac{x \left(0.1 + x\right)}{\left(0.9 - x\right)} = {10}^{- 5} \text{ } \textcolor{red}{\left(3\right)}}$

At this point I am going to make the assumption that, because the value of $\textsf{{K}_{a}}$ is so small, the value of $\textsf{x}$ is negligible compared to $\textsf{0.1}$ and $\textsf{0.9}$.

So we can say:

$\textsf{\left(0.1 + x\right) \Rightarrow 0.1}$

and

$\textsf{\left(0.9 - x\right) \Rightarrow 0.9}$

So sf(color(red)((3)) becomes:

$\textsf{\frac{0.1 x}{0.9} = {10}^{- 5}}$

$\therefore$$\textsf{x = {10}^{- 5} \times \frac{0.9}{0.1} = 9 \times {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l} = \left[{H}^{+}\right]}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left(9 \times {10}^{- 5}\right) = 4.04}$

This derivation is given to explain how to justify the assumptions that are made.

If the value of $\textsf{{K}_{a}}$ is small, as is the case here, then the initial concentrations of $\textsf{H A}$ and $\textsf{{A}^{-}}$ are a close approximation to the equilibrium concentrations.

In practice, there is no need to set up an ICE table and you can go straight to equation sf(color(red)((2)).

You can use moles instead of concentrations as the volume is common to both acid and co - base so cancels.

However, it is a good thing to state the assumptions which I have pointed out as well as the fact that I have ignored the tiny amount of $\textsf{{H}^{+}}$ ions which arise from the auto - ionisation of water.