# Consider the titration of 50.0 mL of 0.20 M NH_3 Kb 1.8x10^-5 with 0.20 M HNO_3. How do you calculate the pH after addition of 50.0mL of the titrant?

Aug 21, 2016

You can do it like this:

#### Explanation:

The equation for the neutralisation is:

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

We can find the number of moles using $\textsf{n = c \times v}$:

$\textsf{{n}_{N {H}_{3}} = 0.20 \times \frac{50.0}{1000} = 0.01}$

$\textsf{{n}_{{H}^{+}} = 0.20 \times \frac{50.0}{1000} = 0.01}$

This tells us that:

$\textsf{{n}_{N {H}_{4}^{+}} = 0.01}$

$\textsf{c = \frac{n}{v}}$

The total volume is 50.0 + 50.0 = 100.0 ml

$\therefore$$\textsf{\left[N {H}_{4}^{+}\right] = \frac{0.01}{\frac{100.0}{1000}} = 0.10 \textcolor{w h i t e}{x} \text{mol/l}}$

The ammonium ion partially dissociates:

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

For which:

sf(K_a=([NH_3][H^+])/([NH_4^+])

To find the $\textsf{p H}$ we need to know the value of $\textsf{{K}_{a}}$. We can get this by using the expression:

$\textsf{{K}_{a} \times {K}_{b} = {K}_{w} = {10}^{- 14} \textcolor{w h i t e}{x} {\text{mol"^2."l}}^{-} 2}$ at $\textsf{{25}^{\circ} C}$

$\therefore$$\textsf{{K}_{a} = {10}^{- 14} / \left(1.8 \times {10}^{- 5}\right) = 5.55 \times {10}^{- 10} \textcolor{w h i t e}{x} \text{mol/l}}$

Now we set up an ICE table based on concentrations:

$\textsf{\text{ "NH_4^+" "rightleftharpoons" "NH_3" "+" } {H}^{+}}$

$\textsf{\textcolor{red}{I} \text{ "0.10" "0" "0}}$

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

$\textsf{\textcolor{red}{E} \text{ "(0.1-x)" "x" } x}$

$\therefore$$\textsf{{K}_{a} = {x}^{2} / \left(0.1 - x\right)}$

Because the value of $\textsf{{K}_{a}}$ is so small we can assume that the size of $\textsf{x}$ is negligible compared with $\textsf{0.1}$ so we can write:

$\textsf{{K}_{a} = {x}^{2} / 0.1 = 5.55 \times {10}^{- 10}}$

$\therefore$$\textsf{{x}^{2} = 5.55 \times {10}^{- 10} \times 0.1 = 5.55 \times {10}^{- 11}}$

$\therefore$$\textsf{x = \left[{H}^{+}\right] = \sqrt{5.55 \times {10}^{- 11}} = 7.449 \times {10}^{- 6} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left[7.449 \times {10}^{- 6}\right] = 5.12}$

$\textsf{\textcolor{red}{p H = 5.12}}$

This shows that the pH is slightly acidic at the equivalence point which is typical of a salt produced from a strong acid and a weak base.