smc-3674-15-pH of Mixed Solution

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.

Show Worked Solution

→ 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%.