smc-3674-15-pH of Mixed Solution

CHEMISTRY, M6 2025 HSC 19 MC

0.1 mol of solid sodium acetate is dissolved in 500 mL of 0.1 mol L$^{-1}$ $\ce{HCl}$ in a beaker. This solution has a pH of 4.8 .

500 mL of distilled water is then added to the beaker.

What is the pH of the final solution?

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$C$

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The initial reaction between sodium acetate and hydrochloric acid runs to completion as $\ce{HCl}$ is a strong acid:
$\ce{CH3COO-(aq) + HCl(aq) -> CH3COOH(aq) + Cl-(aq)}$
$n(\ce{CH3COO-}) = 0.1\ \text{mol}$
$n(\ce{HCl}) = 0.5\ \text{L} \times 0.1\ \text{mol L}^{-1} = 0.05\ \text{mol}$
As they react in a $1:1$ ratio, $\ce{HCl}$ is the limiting reagent.
$n(\ce{CH3COO-_{\text{after reaction}}}) = n(\ce{CH3COO_{\text{initial}}})-n(\ce{CH3COO_{\text{reacted}}}) = 0.1-0.05 = 0.05\ \text{mol}$
$n(\ce{CH3COOH_{\text{after reaction}}}) = n(\ce{CH3COO_{\text{reacted}}}) = 0.05\ \text{mol}$
The following equilibrium reaction is then established below dilution
$\ce{CH3COOH(aq) + H2O(l) \leftrightharpoons CH3COO-(aq) + H3O+(aq)}$
Therefore the following ice table can be constructed:

\begin{array} {|c|c|c|c|}
\hline & \ce{[CH3COOH]} & \ce{[CH3COO-]} & \ce{[H3O+]} \\
\hline \text{Initial} & 0.1 & 0.1 & 0 \\
\hline \text{Change} & -10^{-4.8} & +10^{-4.8} & +10^{-4.8} \\
\hline \text{Equilibrium} & 0.1-10^{-4.8} & 0.1+10^{-4.8} & +10^{-4.8} \\
\hline \end{array}

$\therefore K_a = \dfrac{(0.1+10^{-4.8})(10^{-4.8})}{0.1-10^{-4.8}} = 1.585395 \times 10^{-5}$

Then considering the dilution which would shift the equilibrium position to the right.

\begin{array} {|c|c|c|c|}
\hline & \ce{[CH3COOH]} & \ce{[CH3COO-]} & \ce{[H3O+]} \\
\hline \text{Initial} & 0.1 \times \dfrac{500\ \text{mL}}{1000\ \text{mL}} = 0.05 & 0.1 \times \dfrac{500\ \text{mL}}{1000\ \text{mL}} = 0.05 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 0.05-x & 0.05+x & +x \\
\hline \end{array}

As $K_{eq}$ is small, $0.05 -x \approx 0.05$ and $0.05 + x \approx 0.05$.
$\therefore K_{eq} = \dfrac{0.05x}{0.05} = x = 1.585395 \times 10^{-5}$
$\text{pH} = -\log_{10}\ce{[H3O+]} = -\log_{10}(1.585395 \times 10^{-5}) = 4.79962 = 4.8\ \text{(1 d.p.)}$

$\Rightarrow C$

CHEMISTRY, M6 2024 HSC 29

150 mL of a 0.20 mol L$^{-1}$ sodium hydroxide solution is added to 100 mL of a 0.10 mol L$^{-1}$ sulfuric acid solution.

Calculate the pH of the resulting solution, assuming that the volume of the resulting solution is 250 mL and that its temperature is 25°C. (4 marks)

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$12.60$

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$\ce{H2SO4(aq) + 2NaOH(aq) -> Na2SO4 + 2H2O(l)}$

$\ce{n(NaOH)} = 0.20\ \text{mol L}^{-1} \times 0.150\ \text{L} = 0.03\ \text{mol}$

$\ce{n(H2SO4)} = 0.10\ \text{mol L}^{-1} \times 0.100\ \text{L} = 0.01\ \text{mol}$

$\ce{n(NaOH)}\ \text{required to react with}\ 0.01\ \text{mol of}\ \ce{H2SO4} = 2 \times 0.01 = 0.02\ \text{mol}$.

$\Rightarrow\ \text{Remaining}\ \ce{NaOH} = 0.01\ \text{mol}$.

$\ce{[NaOH]} = \dfrac{0.01\ \text{mol}}{0.250\ \text{L}} = 0.04\ \text{mol L}^{-1} = \ce{[OH^-]}$

$\ce{pOH} = -\log_{10}\ce{[OH-]} = -\log_{10}(0.04) = 1.40$

$\ce{pH} = 14-1.40=12.60$

CHEMISTRY, M6 2008 HSC 14 MC

20 mL 0.08 mol L$^{-1}\ \ce{HCl}$ is mixed with 30 mL of 0.05 mol L$^{-1}\ \ce{NaOH}$.

What is the pH of the resultant solution?

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$B$

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$\ce{n(HCl)_{excess} = 0.0016-0.0015=0.0001 mol}$

\[\ce{[HCl] = \frac{0.0001}{0.020+0.030} = 0.002 mol L^{-1}}\]

$\ce{[HCl] = [H+]}$

$\ce{pH = -log(0.002) = 2.7}$

$\Rightarrow B$

CHEMISTRY, M6 2023 HSC 23

The pH of two solutions, $\text{X}$ and $\text{Y}$, were measured before and after 10 drops of concentrated $ \ce{NaOH} $ was added to each.

Explain the pH changes that occurred in solutions $\text{X}$ and $\text{Y}$. (3 marks)

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The diagram shows that the pH of solution $\text{X}$ changes significantly with the introduction of the base $\ce{NaOH}$, whereas the pH of solution $\text{Y}$ only shows a small change in pH. This indicates that solution $\text{Y}$ contains a buffer while solution $\text{X}$ does not.
When $\ce{NaOH}$ was added to solution $\text{X}$, the addition of $\ce{OH-}$ ions caused the increase in pH $\ce{(pH = −log10 [H3O+]).} $
In contrast, the $\ce{OH-}$ ions react with the buffer solution in solution $\text{Y}$. This has the effect of minimising the change in $\ce{[H3O+]}$ and therefore pH.

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The diagram shows that the pH of solution $\text{X}$ changes significantly with the introduction of the base $\ce{NaOH}$, whereas the pH of solution $\text{Y}$ only shows a small change in pH. This indicates that solution $\text{Y}$ contains a buffer while solution $\text{X}$ does not.
When $\ce{NaOH}$ was added to solution $\text{X}$, the addition of $\ce{OH-}$ ions caused the increase in pH $\ce{(pH = −log10 [H3O+]).} $
In contrast, the $\ce{OH-}$ ions react with the buffer solution in solution $\text{Y}$. This has the effect of minimising the change in $\ce{[H3O+]}$ and therefore pH.

CHEMISTRY, M6 2012 HSC 28

A solution was made by mixing 75.00 mL of 0.120 mol L$^{-1}$ hydrochloric acid with 25.00 mL of 0.200 mol L$^{-1}$ sodium hydroxide.

What is the pH of the solution? (3 marks)

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$\ce{pH = 1.4}$

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$\ce{HCl + NaOH \rightarrow NaCl + H2O}$

$\ce{n(H3O+) = c \times V = 0.120 \times 0.07500 = 0.00900 moles}$

$\ce{n(OH) = c \times V = 0.02500 \times 0.200 = 0.00500 moles}$

$\ce{n(H3O+ excess) = 9 \times 10^{-3}-5 \times 10^{-3} = 4 \times 10^{-3} moles}$

\[\ce{[H3O+] = \frac{4 \times 10^{-3}}{0.100} = 4.00 \times 10^{-2}}\]

$\ce{pH = -log_10[H3O+] = -log_10 4.00 \times 10^{-2} = 1.4}$

CHEMISTRY, M6 EQ-Bank 23

Propanoic acid dissociation in water can is represented in the following equation:

$\ce{CH3CH2COOH($aq$) + H2O($l$) \rightleftharpoons CH3CH2COO^-($aq$) + H3O^{+}($aq$)}$

Explain how the pH of the propanoic acid solution would change if it was diluted. (3 marks)

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Propanoic acid only partially ionises in solution and is defined as a weak acid.
Any dilution of the acid will result in a decrease of the concentration of all species (including the hydronium ion).
According to Le Chatelier, the decreasing concentrations of dissolved species will cause the equilibrium to shift to the right.
While this effect causes an increase in ionisation, it is not sufficient to counter the decrease in hydronium ion concentration caused by the original dilution.
Since the net effect causes the hydronium ion concentration to decrease, the solution will become less acidic and the pH will increase.

Show Worked Solution

Propanoic acid only partially ionises in solution and is defined as a weak acid.
Any dilution of the acid will result in a decrease of the concentration of all species (including the hydronium ion).
According to Le Chatelier, the decreasing concentrations of dissolved species will cause the equilibrium to shift to the right.
While this effect causes an increase in ionisation, it is not sufficient to counter the decrease in hydronium ion concentration caused by the original dilution.
Since the net effect causes the hydronium ion concentration to decrease, the solution will become less acidic and the pH will increase.

CHEMISTRY, M6 EQ-Bank 25

The graph shows changes in pH for the titrations of equal volumes of solutions of two monoprotic acids, Acid 1 and Acid 2.

Explain the differences between Acid 1 and Acid 2 in terms of their relative strengths and concentrations. (3 marks)

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Acid 1 is a strong acid. Its initial pH = 1 and its equivalence point is at pH = 7.
Acid 2 is a weaker acid. Its initial pH ~ 2 and its equivalence point is >7.
Acid 2 has a higher concentration than Acid 1 as it doesn’t take that much more $\ce{KOH}$ to neutralise it.

Show Worked Solution

Acid 1 is a strong acid. Its initial pH = 1 and its equivalence point is at pH = 7.
Acid 2 is a weaker acid. Its initial pH ~ 2 and its equivalence point is >7.
Acid 2 has a higher concentration than Acid 1 as it doesn’t take that much more $\ce{KOH}$ to neutralise it.

CHEMISTRY, M6 2016 HSC 18 MC

40 mL of 0.10 mol L¯¹ $\ce{NaOH}$ is mixed with 60 mL of 0.10 mol L¯¹ $\ce{HCl}$.

What is the pH of the resulting solution?

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`B`

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$\ce{n(HCl)_{excess} = 0.006-0.004=0.002 mol}$

\[\ce{[HCl] = \frac{0.002}{0.040+0.060} = 0.0200 mol L^{-1}}\]

$\ce{[HCl] = [H+]}$

$\ce{pH = -log(0.0200) = 1.7}$

`=> B`

CHEMISTRY, M6 2017 HSC 20 MC

20.0 mL of 0.020 mol L¯¹ barium hydroxide solution is added to 50.0 mL of 0.040 mol L¯¹ hydrochloric acid solution.

What is the pH of the final solution?

`0.2`
`1.6`
`1.8`
`2.9`

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`C`

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$\ce{n(Ba(OH)2) = 0.020 \times 0.02 = 4.00 \times 10^{-4} mol} $

$\ce{n(HCl) = 0.040 \times 0.05 = 2.00 \times 10^{-3} mol} $

$\ce{Ba(OH)_2(aq) + 2HCl(aq) -> BaCl_2(aq) + 2H2O(l)} $

$\ce{Ba(OH)_2}$ is limiting → $\ce{HCl}$ is in excess

$\ce{n(HCl)_{excess} = 2.00 \times 10^{-3} – 2(4.00 \times 10^{-4}) = 1.2 \times 10^{-3} mol} $

\[\ce{[H+] = \frac{n}{total V} = \frac{1.20 \times 10^{-3}}{0.02 + 0.05} = 1.714 \times 10^{-2} mol L^{-1}} \]

$\ce{pH = -log10[H+] = -log10(1.714 \times 10^{-2})} = 1.8$

`=>C`

♦♦ Mean mark 33%.

CHEMISTRY, M6 2021 HSC 15 MC

What is the pH of the resultant solution after 20.0 mL of 0.20 mol `text{L}^{-1}\ text{HCl} (aq)` is mixed with 20.0 mL of 0.50 mol `text{L}^{-1}\ text{NaOH} (aq)`?

11.8
13.2
13.5
14.0

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`B`

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The reaction when `text{HCl} (aq)` is mixed with `text{NaOH} (aq)` is:

`text{HCl}(aq) + text{NaOH}(aq)` ⟶ `text{NaCl} (aq) + text{H}_2 text{O} (aq)`

`text{n(}text{HCl})=\ text{c × V}= 0.20 xx 0.020= 0.0040\ text{mol}`

`text{n(} text{NaOH})=\ text{c × V}=0.50 xx 0.020= 0.010\ text{mol}`

`text{HCl}` is the limiting reagent and `text{NaOH}` is the excess reagent.

`text{n(} text{NaOH) leftover} = 0.010-0.0040 = 0.0060\ text{mol}`

`text{c(} text{NaOH)}= (n(text{NaOH})) / text{V}= 0.0060/0.040=0.15\ text{mol L}^(-1)`

`text{NaOH} (aq)\ ⟶\ text{Na}^(+) + text{OH}^(-)`

Therefore,

`[text{NaOH}] = [text{OH}^(-)] = 0.15\ text{mol L}^(-1)`

`text{pOH} = -text{log}_(10) [text{OH}^(-1)] = -text{log}_(10) (0.15) = 0.8239`

`text{pH}=14- text{pOH}=14-0.8239= 13.2`

`=> B`

♦ Mean mark 54%.