# Question a643f

Nov 19, 2016

$\textsf{0.170 \textcolor{w h i t e}{x} g}$

#### Explanation:

First up I will calculate the pH of the buffer before any HCl is added.

Ammonium ions are very weakly acidic and hydrolyse:

$\textsf{N {H}_{4}^{+} r i g h t \le f t h a r p \infty n s N {H}_{3} + {H}^{+}}$

For which:

$\textsf{{K}_{a} = \frac{\left[N {H}_{3}\right] \left[{H}^{+}\right]}{\left[N {H}_{4}^{+}\right]} = 5.6 \times {10}^{- 10} \textcolor{w h i t e}{x} \text{mol/l"" ""at"" } {25}^{\circ} C}$

These are equilibrium concentrations. To find the pH we need to find $\textsf{\left[{H}^{+}\right]}$ so rearranging:

$\textsf{\left[{H}^{+}\right] = {K}_{a} \times \frac{\left[N {H}_{4}^{+}\right]}{\left[N {H}_{3}\right]} \text{ } \textcolor{red}{\left(1\right)}}$

$\therefore$$\textsf{\left[{H}^{+}\right] = 5.6 \times {10}^{- 10} \times \frac{0.270}{0.390} = 3.877 \times {10}^{- 10} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left(3.877 \times {10}^{- 10}\right) = 9.41}$

Now we need to find how much HCl can be added to bring the pH down to 9.

The buffer works because there is sufficient base to absorb the addition of small amounts of $\textsf{{H}^{+}}$ ions:

$\textsf{{H}^{+} + N {H}_{3} \rightarrow N {H}_{4}^{+}}$

If $\textsf{x}$ moles of $\textsf{{H}^{+}}$ is added then the number of moles of $\textsf{N {H}_{3}}$ goes down by $\textsf{x}$ and the number of moles of $\textsf{N {H}_{4}^{+}}$ goes up by $\textsf{x}$.

If you look at sf(color(red)((1)) you can see that it is the ratio of $\textsf{\frac{\left[N {H}_{4}^{+}\right]}{\left[N {H}_{3}\right]}}$ which can be altered to adjust the pH.

I will adjust sf(color(red)((1))# so it gives a pH of 9 and then find the new ratio of $\textsf{\frac{\left[N {H}_{4}^{+}\right]}{\left[N {H}_{3}\right]}}$. This will allow $\textsf{x}$ to be calculated.

$\textsf{p H = 9}$ so $\textsf{\left[{H}^{+}\right] = {10}^{- p H} = {10}^{- 9} \textcolor{w h i t e}{x} \text{mol/l}}$

$\therefore$$\textsf{{10}^{- 9} = 5.6 \times {10}^{- 10} \times \frac{\left[N {H}_{4}^{+}\right]}{\left[N {H}_{3}\right]}}$.

$\therefore$$\textsf{\frac{\left[N {H}_{4}^{+}\right]}{\left[N {H}_{3}\right]} = 1.785}$

Now to find the new concentrations:

Initial moles of $\textsf{\left[N {H}_{4}^{+}\right] = c \times v = 0.390 \times \frac{130.0}{1000} = 0.507}$

If $\textsf{x}$ moles of $\textsf{{H}^{+}}$ is added then the new moles of $\textsf{N {H}_{4}^{+} \Rightarrow \left(0.0507 + x\right)}$

Initial moles of $\textsf{N {H}_{3} = c \times v = 0.270 \times \frac{130.0}{1000} = 0.0351}$

After adding $\textsf{x}$ moles of $\textsf{{H}^{+}}$ then the new moles of $\textsf{N {H}_{3} \Rightarrow \left(0.351 - x\right)}$

We know the required ratio to give a pH of 9 is 1.785 so we can write:

$\textsf{\frac{\left(0.0507 + x\right)}{\left(0.0357 - x\right)} = 1.785 \text{ } \textcolor{red}{\left(2\right)}}$

Note I can use moles directly in this ratio rather than concentrations as the volume is common to both so will cancel.

Rearranging $\textsf{\textcolor{red}{\left(2\right)} \Rightarrow}$

$\textsf{\left(0.0507 + x\right) = 1.785 \left(0.0357 - x\right)}$

From which:

$\textsf{x = 0.00467 \textcolor{w h i t e}{x} \text{mol}}$

This is the no. of moles of HCl which must be added.

Now we can find the mass of HCl needed:

$\textsf{\text{mass} = n \times {M}_{r} = 0.00467 \times 36.461 = 0.170 \textcolor{w h i t e}{x} g}$

HCl is a gas at room temperature so it would be more practical to add this as a dilute solution of the appropriate concentration and volume.

You can see it is the ratio of acid to co - base concentration which decides the pH.

It is the absolute values of acid and co - base concentration which dictates how much acid or alkali can be dealt with.

This is a function of the "Buffer Capacity".