# Question c25b3

Dec 2, 2016

Approximately 0.04

#### Explanation:

Here's my reasoning:

The 1st ionisation is:

$\textsf{{H}_{3} {A}^{2 +} r i g h t \le f t h a r p \infty n s {H}_{2} {A}^{+} + {H}^{+}}$

For which:

$\textsf{{K}_{a 1} = \frac{\left[{H}_{2} {A}^{+}\right] \left[{H}^{+}\right]}{\left[{H}_{3} {A}^{2 +}\right]}}$

You can see that as the pH rises when alkali is added there comes a point when $\textsf{\left[{H}_{2} {A}^{+}\right] = \left[{H}_{3} {A}^{2 +}\right]}$. The expression becomes:

$\textsf{{K}_{a 1} = \frac{\cancel{\left[{H}_{2} {A}^{+}\right]} \left[{H}^{+}\right]}{\cancel{\left[{H}_{3} {A}^{2 +}\right]}}}$

So

$\textsf{{K}_{a 1} = \left[{H}^{+}\right]}$

And

$\textsf{p {K}_{a 1} = p H}$

This tells us that at pH 3.5 the acid is 1/2 neutralised and we will have an equlimolar mixture of $\textsf{{H}_{3} {A}^{2 +}}$ and $\textsf{{H}_{2} {A}^{+}}$.

At the 1st equivalence point the solution will contain $\textsf{{H}_{2} {A}^{+}}$

The 2nd ionisation is:

$\textsf{{H}_{2} {A}^{+} r i g h t \le f t h a r p \infty n s H A + {H}^{+}}$

By the same reasoning when $\textsf{p {K}_{a 2} = p H = 6.0}$ there will be an equilmolar mixture of $\textsf{{H}_{2} {A}^{+}}$ and $\textsf{H A}$.

We would expect the proportion of $\textsf{{H}_{2} {A}^{+}}$ to $\textsf{H A}$ to decrease as the pH is raised.

The expression for $\textsf{{K}_{a 2}}$ is:

$\textsf{{K}_{a 2} = \frac{\left[H A\right] \left[{H}^{+}\right]}{\left[{H}_{2} {A}^{+}\right]}}$

Rearranging:

$\textsf{\left[{H}^{+}\right] = {K}_{a 2} \times \frac{\left[{H}_{2} {A}^{+}\right]}{\left[H A\right]}}$

Taking -ve logs of both sides gives:

sf(pH=pK_(a2)-log([[H_2A^(+)]]/([HA]))

$\therefore$sf(7.4=6.0-log([[H_2A^(+)]]/[[HA]))#

$\therefore$$\textsf{\log \left(\frac{\left[{H}_{2} {A}^{+}\right]}{\left[H A\right]}\right) = 6.0 - 7.4 = - 1.4}$

From which:

$\textsf{\frac{\left[{H}_{2} {A}^{+}\right]}{\left[H A\right]} = 0.0398}$

This means the solution is about 4% $\textsf{{H}_{2} {A}^{+}}$.

I'm not sure what is meant by "the average charge" so suggest it would be about 0.04.