# Question #eab13

Aug 16, 2016

$\textsf{p H = 4.1}$

#### Explanation:

I got a different value for the initial pH. Here's what I got:

Benzoic acid is a weak acid and dissociates:

$\textsf{{C}_{6} {H}_{5} C O O H r i g h t \le f t h a r p \infty n s {C}_{6} {H}_{5} C O {O}^{-} + {H}^{+}}$

For which:

$\textsf{{K}_{a} = \frac{\left[{C}_{6} {H}_{5} C O {O}^{-}\right] \left[{H}^{+}\right]}{\left[{C}_{6} {H}_{5} C O O H\right]}}$

Rearranging:

$\textsf{\left[{H}^{+}\right] = {K}_{a} \times \frac{\left[{C}_{6} {H}_{5} C O O H\right]}{\left[{C}_{6} {H}_{5} C O {O}^{-}\right]}}$

Since $\textsf{p {K}_{a} = 4.2}$ this means $\textsf{{K}_{a} = 6.31 \times {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

I will assume that the no. of moles given refers to equilibrium moles. Since the volume is 1L we can write:

$\textsf{\left[{H}^{+}\right] = 6.31 \times {10}^{- 5} \times \frac{0.3}{0.35} = 5.4 \times {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left(5.4 \times {10}^{- 5}\right) = 4.26}$

Now a small amount of HCl is added. We have a buffer solution here with a large reserve of benzoate ions which can absorb the addition of small amounts of $\textsf{{H}^{+}}$ ions :

$\textsf{{C}_{6} {H}_{5} C O {O}^{-} + {H}^{+} \rightarrow {C}_{6} {H}_{5} C O O H}$

If we add 0.05 moles of $\textsf{{H}^{+}}$ ions you can see that 0.05 moles of benzoate ions will be consumed and 0.05 moles of benzoic acid will be formed.

So we can say:

$\textsf{{n}_{{C}_{6} {H}_{5} C O {O}^{-}} = 0.35 - 0.05 = 0.3}$

and

$\textsf{{n}_{{C}_{6} {H}_{5} C O O H} = 0.3 + 0.05 = 0.35}$

There may be a volume change on mixing but this does not matter as the volume is common to acid and co - base so cancels.

We can now put these values into our original expression for $\textsf{\left[{H}^{+}\right] \Rightarrow}$

$\textsf{\left[{H}^{+}\right] = 6.31 \times {10}^{- 5} \times \frac{0.35}{0.3} = 7.36 \times {10}^{- 5} \textcolor{w h i t e}{x} \text{mol/l}}$

$\therefore$$\textsf{p H = - \log \left(7.36 \times {10}^{- 5}\right) = 4.1}$

You can see that the buffer has worked since the pH has fallen only very slightly.