# Question #d77d9

Jan 13, 2017

$\textsf{p H = 9}$

#### Explanation:

I will you assume you mean 20 ml and not 20 L. That would be a ridiculous volume for a titration.

Let the general equation be:

$\textsf{H X + Y O H \rightarrow X Y + {H}_{2} O}$

This tells us that 1 mole of HX produces 1 mole of XY.

The number of moles of HX is given by:

$\textsf{{n}_{H X} = c \times v = 0.2 \times \frac{20}{1000} = 0.004}$

$\therefore$$\textsf{{n}_{X Y} = 0.004}$

This is the number of moles of XY at the equivalence point.

XY is the salt of a weak acid and a strong base. The solution formed will not be neutral because the $\textsf{{Y}^{-}}$ ion is basic and undergoes hydrolysis:

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

By setting up an ICE table you can obtain a useful expression to get the pOH and hence the pH:

$\textsf{p O H = \frac{1}{2} \left[p {K}_{b} - \log b\right]}$

Where b is the concentration of the base i.e $\textsf{\left[{Y}^{-}\right]}$

Note that the total volume after the titration = 20 ml + 20 ml = 40 ml

$\therefore$$\textsf{\left[{Y}^{-}\right] = \frac{n}{v} = \frac{0.004}{\frac{40}{1000}} = 0.1 \textcolor{w h i t e}{x} \text{mol/l}}$

To get $\textsf{p {K}_{b}}$ we use:

$\textsf{p {K}_{a} + p {K}_{b} = p {K}_{w} = 14}$ at $\textsf{{25}^{\circ} C}$

$\therefore$$\textsf{p {K}_{b} = 14 - p {K}_{a} = 14 - 5 = 9}$

$\therefore$$\textsf{p O H = \frac{1}{2} \left[9 - \log \left(0.1\right)\right] = \frac{1}{2} \left[10\right] = 5}$

We know that:

$\textsf{p H + p O H = p {K}_{w} = 14}$

$\therefore$$\textsf{p H = 14 - p O H = 14 - 5 = 9}$

We would expect a result like this as the solution should be alkaline.