# A 100.0 mL sample of 0.10 M NH_3 is titrated with 0.10 M HNO_3. How do you determine the pH of the solution after the addition of 40.0 mL of HNO_3?. The Kb of NH3 is 1.8 × 10-5.?

Dec 29, 2016

You can do it like this:

#### Explanation:

As the acid is added to the base the following neutralisation takes place:

$\textsf{N {H}_{3 \left(a q\right)} + H N {O}_{3 \left(a q\right)} \rightarrow N {H}_{4} N {O}_{3 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)}}$

The initial moles of $\textsf{N {H}_{3}}$ present is given by:

$\textsf{{n}_{N {H}_{3}} = c \times v = 0.10 \times \frac{100.0}{1000} = 0.01}$

The number of moles of $\textsf{H N {O}_{3}}$ added is given by:

$\textsf{{n}_{H N {O}_{3}} = c \times v = 0.10 \times \frac{40.0}{1000} = 0.004}$

From the equation you can see that the acid and base react in a molar ratio of 1:1.

So the no. moles of $\textsf{N {H}_{3}}$ remaining will be 0.01 - 0.004 = 0.006.

The total volume is now $\textsf{100.0 + 40.0 = 140.0 \textcolor{w h i t e}{x} c {m}^{3}}$

The concentration of $\textsf{N {H}_{3}}$ is given by:

$\textsf{\left[N {H}_{3}\right] = \frac{c}{v} = \frac{0.006}{\frac{140.0}{1000}} = 0.04286 \textcolor{w h i t e}{x} \text{mol/l}}$

From an ICE table we get the expression:

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

Where b is the concentration of the base.

We can approximate this to the initial concentration since the dissociation is small.

$\textsf{p {K}_{b} = - \log {K}_{b} = - \log \left(1.8 \times {10}^{- 5}\right) = 4.744}$

Putting in the numbers:

sf(pOH=1/2[4.744-log(0.04286)]

$\textsf{p O H = \frac{1}{2} \left[4.744 - \left(- 1.3679\right)\right] = 3.056}$

At $\textsf{{25}^{\circ} \textcolor{w h i t e}{x} C}$ we know that:

$\textsf{p H + p O H = 14}$

$\therefore$$\textsf{p H = 14 - 3.056 = 10.9}$