If a buffer solution is prepared from 5.15g NH_4NO_3 and 0.10L of 0.15M NH_3, what is the pH of the solution?

Jun 23, 2016

$p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

Explanation:

It can be shown relatively easily that the $p H$ of a weak acid mixed with its conjugate base in appreciable quantities conforms to the so-called $\text{buffer equation}$ as given.

You have not quoted $p {K}_{a}$ for ammonium ion, but I will take pity on you, because I know that $p {K}_{a}$, $N {H}_{4}^{+} = 9.24$. Note that you are not expected to know these $p {K}_{a}$ values from memory even as an undergraduate. You ARE expected to know that ammonium ion is a fairly weak acid, and how to manipulate the buffer equation.

Given the $p {K}_{a}$ it is simply a question of determining the concentrations and plugging in the numbers:

$p H = 9.24 + {\log}_{10} \left\{\frac{0.015 \cdot m o l \cdot {L}^{-} 1}{\frac{5.15 \cdot g}{80.04 \cdot \frac{g}{m o l}} \times \frac{1}{0.10 \cdot L}}\right\}$ $=$ ??

Note that here I knew that volume of the solution was $0.10 \cdot L$. How did I know this?

PS I hope you can see the units in my calculations, because I can't! It is important that the dimensions in the log argument cancel out, which I think they do!

Jun 23, 2016

$\text{pH=8.62}$

Explanation:

The ammonium ion is a weak acid and dissociates:

$N {H}_{4 \left(a q\right)}^{+} r i g h t \le f t h a r p \infty n s N {H}_{3 \left(a q\right)} + {H}_{\left(a q\right)}^{+}$

For which:

${K}_{a} = \frac{\left[N {H}_{3 \left(a q\right)}\right] \left[{H}_{\left(a q\right)}^{+}\right]}{\left[N {H}_{4 \left(a q\right)}^{+}\right]} = 5.6 \times {10}^{- 10} \text{ ""mol/l"" }$ at ${25}^{\circ} \text{C}$

To find the $\text{pH}$ we need to calculate the concentration of the ${H}^{+}$ ions.

Rearranging gives:

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

These are all equilibrium concentrations. We can set up an "ICE"  table to find these.

For the initial moles of $N {H}_{3}$ we use $n = c \times v$:

$\therefore n N {H}_{3 \left(a q\right)} = 0.15 \times 0.1 = 0.015$

The initial moles of $N {H}_{4} N {O}_{3}$ is equal to $\frac{m}{M} _ r = \frac{5.15}{80.04} = 0.0643$.

From the formula we can see that the initial moles of $N {H}_{4 \left(a q\right)}^{+}$ must also be $0.0643$.

Now we can set up the $\text{ICE}$ table:

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

$\textcolor{red}{\text{I"" "0.0643" "0.015" }} 0$

color(red)("C")" "-x" "+x" "+x

color(red)("E")" "(0.0643-x)" "(0.015+x)" "x

Because the value of ${K}_{a}$ is so small the value of $x$ is much smaller than $0.0643$ and $0.015$ so we can assume that:

$\left(0.0643 - x\right) \Rightarrow 0.0643$

and

$\left(0.015 + x\right) \Rightarrow 0.015$

Now we can put the numbers into $\textcolor{red}{\left(1\right)} \Rightarrow$

$\left[{H}_{\left(a q\right)}^{+}\right] = 5.6 \times {10}^{- 10} \times \frac{0.0643}{0.015} = 2.4 \times {10}^{- 9} \text{ ""mol/l}$

You will note I can use moles and not concentrations as the volume is common to both $\left[N {H}_{4}^{+}\right]$ and $\left[N {H}_{3}\right]$ so cancels.

This also accounts for any volume change when the solid salt is added to the solution.

$p H = - \log \left[{H}_{\left(a q\right)}^{+}\right]$

$\therefore p H = - \log \left[2.4 \times {10}^{- 9}\right] = \textcolor{red}{8.62}$

In practice you could go straight to $\textcolor{red}{\left(1\right)}$ but bear in mind the assumptions you have made in approximating equilibrium concentrations to initial ones.

This only works if ${K}_{a}$ is very small. If not, then you need to use an $\text{ICE}$ table.