# A solution P contains 0.55 mol/L ammonia and 0.10 mol/L ammonium chloride. What is the pH of the solution after the addition of 1.0 mL of 0.10 mol/L NaOH?

Feb 20, 2016

See explanation.

#### Explanation:

This answer is incorrect because it does not take into account the fact that you're dealing with a basic buffer solution.

The trick here was to recognize the fact that the solution contains ammonium hydroxide, which is actually a solution of ammonia, ${\text{NH}}_{3}$, a weak base, and ammonium chloride, the salt of ammonia's conjugate acid, the ammonium ion, ${\text{NH}}_{4}^{+}$.

A buffer solution is able to resist changes to its pH upon the addition of small amounts of strong acid or strong base.

This means that, even without doing any calculation, you should have recognized that the pH of the solution will virtually remain unchanged by the addition of the sodium hydroxide solution.

Correct answer: $\text{pH} = 10.0$

Since I didn't recognize the fact that this was a buffer solution, I will leave my inaccurate answer below to serve as an example of how NOT to answer this question.

$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a a a a a a a a a a a}} \textcolor{red}{\text{Incorrect answer below}} \frac{\textcolor{w h i t e}{a a a a a a a a a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a}}$

The trick here is that the pH of the solution is the only information you actually need here.

Don't worry about what solution $\text{P}$ is said to contain, that is not important as long as the pH of the solution is said to be equal to $10.0$.

First thing first, try to predict if you expect the pH of the solution to increase or to decrease upon the addition of the sodium hydroxide, $\text{NaOH}$, solution.

Well, since sodium hydroxide is a strong base, it will dissociate completely to form sodium cations, which are of no interest to you, and hydroxide anions, ${\text{OH}}^{-}$.

So, adding the sodium hydroxide will Increase the concentration of the hydroxide anions, which can only mean that the pH of the solution will increase, i.e. the solution will become even more basic.

Prediction:

$\text{pH} > 10.0$

Use the pH of the solution to find the concentration of the hydroxide anions before adding the sodium hydroxide

$\textcolor{b l u e}{\text{pH" + "pOH} = 14}$

Therefore, you have

$\text{pOH} = 14 - 10.0 = 4.0$

This means that the concentration of hydroxide anions is

"pOH"= - log(["OH"^(-)]) implies ["OH"^(-)] = 10^(-"pOH")

["OH"^(-)] = 10^(-4)"M"

Now, you take a ${\text{1-dm}}^{3}$ sample of this solution $\text{P}$. Use the volume and the molarity of the hydroxide anions to calculate how many moles you have present

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${n}_{O {H}^{-}} = {10}^{- 4} {\text{mol" color(red)(cancel(color(black)("dm"^(-3)))) * 1 color(red)(cancel(color(black)("dm"^3))) = 10^(-4)"moles OH}}^{-}$

Now use the molarity and volume of the sodium hydroxide solution to determine how many moles of hydroxide anions you're adding to that sample of solution $\text{P}$

n_(OH^(-)"added") = "0.1 mol" color(red)(cancel(color(black)("dm"^(-3)))) * 1 * 10^(-3)color(red)(cancel(color(black)("dm"^3))) = 10^(-4)"mol OH"^(-)

Now, the total number of moles of hydroxide anions will be

${n}_{O {H}^{-} \text{total") = n_(OH^(-)) + n_(OH^(-)"added}}$

n_(OH^(-)"total") = 10^(-4)"moles" + 10^(-4)"moles" = 2 * 10^(-4)"moles OH"^(-)

The total volume of the resulting solution will be

${V}_{\text{total" = V_"sample P" + V_"added}}$

${V}_{\text{total" = "1 dm"^3 + 1 * 10^(-3)"dm"^3 = "1.001 dm}}^{3}$

The molarity of the hydroxide anions after you add the sodium hydroxide solution will be

["OH"^(-)] = (2 * 10^(-4)"moles")/"1.001 dm"^3 = 1.998 * 10^(-4)"M"

The pOH of the resulting solution will be

$\text{pOH} = - \log \left(1.998 \cdot {10}^{- 4}\right) = 3.70$

The pH of this solution will be

$\text{pH} = 14 - 3.70 = \textcolor{g r e e n}{10.3}$

As predicted, the pH of the solution increased upon the addition of the sodium hydroxide solution.

Feb 21, 2016

$p H = 10$

i.e unchanged.

#### Explanation:

This is a recipe for an alkaline buffer. It resists the addition of small amounts of acid and alkali keeping the pH constant.

How does it work?

You need a solution of a weak base and its salt. In this case we have ammonia solution and the salt ammonium chloride.

$N {H}_{4} O H$ should properly be described as dilute ammonia solution. It forms an alkaline solution in water:

$N {H}_{3 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s N {H}_{4 \left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-} \text{ } \textcolor{red}{\left(1\right)}$

Ammonium chloride is an ionic salt and dissociates completely in water:

$N {H}_{4} C {l}_{\left(s\right)} \rightarrow N {H}_{4 \left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-} \text{ } \textcolor{red}{\left(2\right)}$

If a small amount of ${H}^{+}$ ions is added then they will react immediately react with the $O {H}^{-}$ ions from $\textcolor{red}{\left(1\right)}$. The position of equilibrium will then shift to the right to restore the $O {H}^{-}$ ions thus restoring the pH value.

In this question a small amount of $O {H}^{-}$ ions is added. From $\textcolor{red}{\left(2\right)}$ you can see that there is a large reserve of $N {H}_{4}^{+}$ ions present. These can combine with the $O {H}^{-}$ thus driving $\textcolor{red}{\left(1\right)}$ to the left and restoring the pH value.

A calculation can show this:

Ammonium ions dissociate:

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

For which:

${K}_{a} = \frac{\left[N {H}_{3}\right] \left[{H}^{+}\right]}{\left[N {H}_{4}^{+}\right]} = 5.62 \times {10}^{- 10} \text{mol/l}$ at 298K $\text{ } \textcolor{red}{\left(3\right)}$

The number of moles of $O {H}^{-}$ ions added $= 1 \times \frac{0.1}{1000} = 0.0001$

These will combine with the $N {H}_{4}^{+}$ ions:

$N {H}_{4}^{+} + O {H}^{-} \rightarrow N {H}_{3} + {H}_{2} O$

In 1 litre of P we have 0.1 mol of $N {H}_{4}^{+}$ ions. (I have ignored the small amount from the dissociation of $N {H}_{3}$).

So we are left with $0.1 - 0.0001 = 0.0999$ moles of $N {H}_{4}^{+}$

By the same reasoning you can see that the number of moles of ammonia must have increased by this amount.

This gives $0.55 + 0.0001 = 0.5501$ moles of $N {H}_{3}$.

To find the pH of a buffer we rearrange $\textcolor{red}{\left(3\right)} \Rightarrow$

$\left[{H}^{+}\right] = {K}_{a} \times \frac{\left[N {H}_{4}^{+}\right]}{\left[N {H}_{3}\right]}$

Concentration = moles/volume. We can ignore the total volume of the solution since it is common to both $\left[N {H}_{4}^{+}\right]$ and $\left[N {H}_{3}\right]$.

$\therefore \left[{H}^{+}\right] = {K}_{a} \times \frac{0.0999}{0.5501}$

$\left[{H}^{+}\right] = 5.62 \times {10}^{- 10} \times 0.181 = 1.02 \times {10}^{- 10}$

$\therefore p H = - \log \left(1.02 \times {10}^{- 10}\right) = 10$

So the pH does not change.

This is for illustration only. As you state, this is a multi - choice question so my take on this is that you are required recognise it is the recipe for an alkali buffer so reason that the pH won't change.