# Consider the titration of 300.0 mL of 0.300 MNH_3 (K_b = 1.8 xx 10^-5) with 0.300 M HNO_3. What is the pH at the equivalence point?

Jul 11, 2016

$\textsf{p H = 5.15}$

#### Explanation:

At the equivalence point the reaction is complete:

sf(NH_(3(aq))+HNO_(3(aq))rarrNH_4NO_(3(aq))

The number of moles of $\textsf{N {H}_{3}}$ reacting = $\textsf{0.3 \times \frac{300}{1000} = 0.09}$

So the number of moles of $\textsf{N {H}_{4}^{+}}$ formed = $\textsf{0.09}$

Now we can set up an ICE table:

$\text{ }$$\textsf{N {H}_{4}^{+} \text{ "rightleftharpoons" "NH_3" "+" } {H}^{+}}$

$\textsf{\textcolor{red}{I} \text{ " 0.09" "0" } 0}$

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

$\textsf{\textcolor{red}{E} \text{ "(0.09-x)" "x" } x}$

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

We can find $\textsf{{K}_{a}}$ using:

$\textsf{{K}_{w} = {K}_{a} \times {K}_{b}}$

$\therefore$$\textsf{{K}_{a} = \frac{{10}^{- 14}}{1.8 \times {10}^{- 5}} = 5.55 \times {10}^{- 10} \text{ ""mol/l}}$ at $\textsf{{25}^{\circ} C}$

The concentrations are at equilibrium so we can write:

$\textsf{{K}_{a} = \frac{x . x}{\left(0.09 - x\right)} = 5.55 \times {10}^{- 10}}$

Because the value of $\textsf{{K}_{a}}$ is so small we can assume that $\textsf{\left(0.09 - x\right) \Rightarrow 0.9}$.

$\therefore$$\textsf{\frac{{x}^{2}}{0.09} = 5.55 \times {10}^{- 10}}$

$\therefore$$\textsf{{x}^{2} = 4.99 \times {10}^{- 11}}$

$\textsf{x = 7.07 \times {10}^{- 6} = {\left[{H}^{+}\right]}_{e q m}}$

$\textsf{p H = - \log {\left[{H}^{+}\right]}_{e q m} = - \log \left[7.07 \times {10}^{- 6}\right]}$

$\textsf{p H = 5.15}$

This is an example of "salt hydrolysis". When a strong acid and a weak base produce a salt the final product is slightly acidic.