smc-3674-10-Calculations Involving pH

CHEMISTRY, M6 2024 HSC 28

Iodic acid and sulfamic acid are monoprotic acids. A 0.100 mol L$^{-1}$ solution of iodic acid has a pH of 1.151, as does a 0.120 mol L$^{-1}$ solution of sulfamic acid.

Show that neither iodic acid nor sulfamic acid dissociates completely in water, and determine which is the stronger acid. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

→ Calculating the concentration of hydronium ions in solution for a pH of 1.151.

$\ce{pH}$	$=-\log_{10}\ce{[H3O+]}$
$\ce{[H3O+]}$	$=10^{-\ce{pH}}=10^{-1.151}=0.0706$

→ As this is less than the concentration of both of the acids, neither acid completely dissociates in water.

→ A smaller concentration of iodic acid (0.100 mol/L compared to 0.120 mol/L sulfamic acid) produces the same pH level. Iodic acid must have a greater extent of ionisation compared to sulfamic acid.

→ Therefore iodic acid is a stronger acid than sulfamic acid.

Show Worked Solution

→ Calculating the concentration of hydronium ions in solution for a pH of 1.151.

$\ce{pH}$	$=-\log_{10}\ce{[H3O+]}$
$\ce{[H3O+]}$	$=10^{-\ce{pH}}=10^{-1.151}=0.0706$

→ As this is less than the concentration of both of the acids, neither acid completely dissociates in water.

→ Therefore iodic acid is a stronger acid than sulfamic acid.

CHEMISTRY, M6 2024 HSC 6 MC

What is the hydroxide ion concentration of a solution of potassium hydroxide with a pH of 11 ?

$10^{-11} \text{ mol L}^{-1}$
$10^{-3} \text{ mol L}^{-1}$
$10^3 \text{ mol L}^{-1}$
$10^{11}\text{ mol L}^{-1}$

Show Answers Only

$B$

Show Worked Solution

$\text{pOH}\ =14-11=3$

$\ce{[OH-]}\ =10^{-\text{pOH}}\ =10^{-3}\ \text{mol L}^{-1}$

$\Rightarrow B$

CHEMISTRY M6 2016 VCE 21*

The ammonium ion $\ce{NH4+}$ acts as a weak acid according to the equation

$\ce{NH4+(aq) + H2O(l) \rightleftharpoons NH3(aq) + H3O+(aq)}$

Given the $K_a \ce{(NH4+) = 5.6 \times 10^{-10}}$, determine the [$\ce{H3O+}$] of a 0.200 M ammonium chloride solution. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

$1.06 \times 10^{-5}\ \text{mol L}^{-1}$

Show Worked Solution

$K_a = \dfrac{\ce{[NH3][H3O+]}}{\ce{[NH4+]}} $

$\text{Weak acid assumptions:}$

$\ce{[NH4+]_{eq} = 0.200\ M\ \ \text{and}\ \ [NH3] = [H3O+]} $

$5.6 \times 10^{-10}$	$= \dfrac{\ce{[H3O+]^2}}{0.200} $
$\ce{[H3O+]^2}$	$=0.200 \times 5.6 \times 10^{-10} $
$\ce{[H3O+]}$	$= \sqrt{1.12 \times 10^{-10}}$
	$=1.06 \times 10^{-5}\ \text{mol L}^{-1} $

CHEMISTRY, M6 2015 VCE 22 MC

What is the pH of a 0.0500 M solution of barium hydroxide, $\ce{Ba(OH)2}$?

1.00
1.30
12.7
13.0

Show Answers Only

$D$

Show Worked Solution

$\ce{Ba(OH)2(aq) \rightarrow Ba^{2+}(aq) + 2OH-(aq)} $

$\ce{[OH-] = 2 \times [Ba(OH)2] = 2 \times 0.0500 = 0.100 M} $

$\ce{[H3O+]} = \dfrac{10^{-14}}{\ce{[OH-]}} = \dfrac{10^{-14}}{0.100} = 10^{-13}\ \text{M} $

$\text{pH}\ = -\log_{10} 10^{-13} = 13$

$\Rightarrow D$

♦ Mean mark 49%.

CHEMISTRY, M6 2015 VCE 8

Hydrogen sulfide, in solution, is a diprotic acid and ionises in two stages.

$\ce{H2S(aq) + H2O(l)\rightleftharpoons HS-(aq) + H3O+(aq)}$ $\quad K_{a1} = 9.6 × 10^{–8} \text{ M}$

$\ce{HS–(aq) + H2O(l)\rightleftharpoons S^{2-}(aq) + H3O+(aq)}$ $\quad K_{a2} = 1.3 × 10^{–14} \text{ M}$

A student made two assumptions when estimating the pH of a $0.01 \text{ M}$ solution of $\ce{H2S}$:

Assumption 1: The pH can be estimated by considering only the first ionisation reaction.

Assumption 2: The concentration of $\ce{H2S}$ at equilibrium is approximately equal to $0.01 \text{ M}$.

Explain why these two assumptions are justified. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Use the two assumptions given above to calculate the pH of a 0.01 M solution of $\ce{H2S}$. (3 marks)

--- 5 WORK AREA LINES (style=lined) ---
Some solid sodium hydrogen sulfide, $\ce{NaHS}$, is added to a 0.01 M solution of $\ce{H2S}$.
Predict the effect of this addition on the pH of the hydrogen sulfide solution. Justify your prediction. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

a. 1st assumption:

→ $K_{a2}$ is significantly smaller than the first ionisation $\ce{(K_{a1})}$, making its impact on the $\ce{[H3O+]}$ / pH level negligible.

2nd assumption:

→ $K_{a1}$ is very small, making the extent of the ionisation of $\ce{H2S}$ very small and hence a minimal change in $\ce{[H2S]}$ results.

b. $\text{pH}\ = -\log{10}(3.1 \times 10^{-5}) = 4.5 $

c. Adding $\ce{NaHS}$:

→ Increases the $\ce{[HS-]}$.

→ This increase causes the 1st ionisation equilibrium back to the left.

→ This left shift in the equilibrium decreases the $\ce{[H3O+]}$ and the pH will therefore increase.

Show Worked Solution

a. 1st assumption:

→ $K_{a2}$ is significantly smaller than the first ionisation $\ce{(K_{a1})}$, making its impact on the $\ce{[H3O+]}$ / pH level negligible.

2nd assumption:

→ $K_{a1}$ is very small, making the extent of the ionisation of $\ce{H2S}$ very small and hence a minimal change in $\ce{[H2S]}$ results.

♦♦ Mean mark (a) 27%.

b.	$K_{a1}$	$=\dfrac{\ce{[HS-][H3O+]}}{\ce{[H2S]}} $
	$9.6 \times 10^{-8}$	$=\dfrac{\ce{[H3O+]^2}}{0.01}$
	$\ce{[H3O+]^2}$	$=0.01 \times 9.6 \times 10^{-8} $
	$\ce{[H3O+]}$	$=\sqrt{9.6 \times 10^{-10}}$
		$=3.1 \times 10^{-5}\ \text{M} $

$\text{pH}\ = -\log{10}(3.1 \times 10^{-5}) = 4.5 $

c. Adding $\ce{NaHS}$:

→ Increases the $\ce{[HS-]}$.

→ This increase causes the 1st ionisation equilibrium back to the left.

→ This left shift in the equilibrium decreases the $\ce{[H3O+]}$ and the pH will therefore increase.

♦♦ Mean mark (c) 35%.

CHEMISTRY, M6 2023 HSC 6 MC

The pH of a solution changes from 8 to 5.

What happens to the concentration of hydrogen ions during this change of pH?

It increases by a factor of 3.
It decreases by a factor of 3.
It increases by a factor of 1000.
It decreases by a factor of 1000.

Show Answers Only

$C$

Show Worked Solution

→ Each increase/decrease of pH by a magnitude of 1 represents a change in $\ce{[H+]}$ of a factor of 10.

→ Therefore, concentration change when pH moves from 8 to 5 = 10 × 3 = 1000 (increase).

$\Rightarrow C$

CHEMISTRY, M6 2018 HSC 18 MC

The pH of a 0.080 mol L¯¹ solution of acetic acid is 2.9.

What percentage of the acetic acid has dissociated into ions?

1.0%
1.3%
1.6%
2.8%

Show Answers Only

`C`

Show Worked Solution

$\ce{pH = -log[H+]}$

$\ce{[H+] = 10^{-pH} = 10^{-2.9} = 1.259 mol L^{-1}}$

`%`

`=1.259/0.08\ times 100`

`=1.6`

`=>C`

CHEMISTRY, M6 2016 HSC 12 MC

Which of the following could be added to 100 mL of 0.01 mol L¯¹ hydrochloric acid solution to change its pH to 4?

900 mL of water
900 mL of 0.01 mol L¯¹ hydrochloric acid
9900 mL of water
9900 mL of 0.01 mol L¯¹ hydrochloric acid

Show Answers Only

`C`

Show Worked Solution

By Elimination:

→ If 0.01 mol L¯¹ hydrochloric acid is added, $\ce{[H+]}$ remains unchanged and hence pH is unchanged (eliminate B and D)

→ If 900 mL of water added, $\ce{[H+]}$ dilutes to 0.001 mol L¯¹, with a pH = 3 (eliminate A)

→ If 9900 mL of water added, $\ce{[H+]}$ dilutes to 0.0001 mol L¯¹, with a pH = 4

`=>C`

Mean mark 59%.

CHEMISTRY, M6 2019 HSC 33

A student adds 1.17 g of $\ce{Al(OH)3 (s)}$ to 0.500 L of 0.100 mol L¯¹ $\ce{HCl(aq)}$.

Calculate the pH of the resulting solution. Assume that the volume of the resulting solution is 0.500 L. (4 marks)

Show Answers Only

$\ce{pH = -log(0.010) = 2.00}$

Show Worked Solution

$\ce{Al(OH)3(s) + 3HCl(aq) -> AlCl3(aq) + 3H2O(l)}$

\[\ce{n(Al(OH)3) = \frac{m}{MM} = \frac{1.17}{78.004} = 0.0150 mol}\]

$\ce{n(HCl) = c \times V = 0.500 \times 0.100 = 0.050 mol}$

→ $\ce{Al(OH)3}$ is limiting reagent $ \rightarrow \ce{HCl}$ is excess reagent

$\ce{n(HCl)_{reacted} = 3 \times 0.015 = 0.0450 mol}$

$\ce{n(HCl)_{excess}}$	$\ce{= n(HCl)_{init}-n(HCl)_{reacted}}$
	`=0.0500-0.0450`
	`=0.005\ \text{mol}`

$\ce{HCl(aq) -> H+(aq) + Cl-(aq)}$

$\ce{n(H+) = n(HCl)_{excess} = 0.005 mol}$

\[\ce{[H+] = \frac{n}{V} = \frac{0.005}{0.5} = 0.010 mol L^{-1}}\]

$\ce{pH = -log[H+] = -log(0.010) = 2.00}$

Mean mark 56%.

CHEMISTRY, M6 2019 HSC 27

The relationship between the acid dissociation constant, `K_a`, and the corresponding conjugate base dissociation constant, `K_b`, is given by:

`K_(a)xxK_(b)=K_(w)`

Assume that the temperature for part (a) and part (b) is 25°C.

The `K_a` of hypochlorous acid `text{(HOCl)}` is `3.0 xx10^(-8)`.
Show that the `K_b` of the hypochlorite ion, `text{OCl}^-`, is `3.3 xx10^(-7)`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
The conjugate base dissociation constant, `K_b`, is the equilibrium constant for the following equation:
`text{OCl}^(-)(aq)+ text{H}_(2) text{O}(l) ⇌ text{HOCl}(aq)+ text{OH}^(-)(aq)`
Calculate the pH of a 0.20 mol L¯¹ solution of sodium hypochlorite `(text{NaOCl})`. (4 mark)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`K_b=3.3 xx 10^{-7}`
$\ce{pH = 14-3.59 = 10.41}$

Show Worked Solution

a. `K_(a)xxK_(b)=K_(w)\ \ =>\ \ K_b=(K_(w))/(K_(a))`

`K_b`	`=(1.0 xx 10^{-14})/(3.0 xx 10^{-8}`
	`=3.3 xx 10^{-7}`

b. $\ce{OCl-(aq) + H2O(l) \rightleftharpoons HOCl(aq) + OH-(aq)}$

\begin{array} {|l|c|c|c|}
\hline & \ce{OCl-} & \ce{HOCl} & \ce{OH–} \\
\hline \text{Initial} & 0.20 & 0 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 0.20-x & x & x \\
\hline \end{array}

\[ K_b = \ce{\frac{[HOCl][OH– ]}{[OCl-]}} = \frac{x^2}{(0.20-x)} \]

Assume `0.20-x~~0.20` because `x` is negligible:

`3.3 xx 10^(-7)`	`= x^2 / (0.20-x)`
`x`	`=sqrt(3.3 xx 10^(−7) xx 0.20)`
	`= 2.5690 xx 10^{-4}\ text{mol L}^(–1)`

$\ce{[OH-] = 2.5690 \times 10^{-4} mol L^{-1}}$

$\ce{pOH = -log10[OH-] = -log10(2.5690 \times 10^{-4}) = 3.59}$

$\therefore \ce{pH = 14-3.59 = 10.41}$

♦ Mean mark (b) 45%.

CHEMISTRY, M6 2019 HSC 15 MC

What is the concentration of hydroxide ions (in mol L ¯¹) in a solution that has a pH of 8.53?

`3.0 xx10^(-9)`
`3.4 xx10^(-6)`
`5.5`
`3.0 xx10^(5)`

Show Answers Only

`B`

Show Worked Solution

`text{pOH}`	`=14-text{pH}`
	`=14-8.53`
	`=5.47`

$\ce{[OH-]}$	`=10^(text{–pH})`
	`=10^(-5.47)`
	`=3.4 xx10^(-6)`

`=>B`

CHEMISTRY, M6 2020 HSC 33

Excess solid calcium hydroxide is added to a beaker containing 0.100 L of 2.00 mol L¯¹ hydrochloric acid and the mixture is allowed to come to equilibrium.

Show that the amount (in mol) of calcium hydroxide that reacts with the hydrochloric acid is 0.100 mol. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
It is valid in this instance to make the simplifying assumption that the amount of calcium ions present at equilibrium is equal to the amount generated in the reaction in part (a).
Calculate the pH of the resulting solution. (4 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $\ce{Ca(OH)2 (s) + 2 HCl (aq) -> CaCl2 (aq) + 2 H2O (l)}$

`text{n(HCl)} = text{c} xx text{V} = 2.00 xx 0.100 = 0.200\ text{mol}`

\[\ce{n(Ca(OH)2) = \frac{n(HCL)}{2} = \frac{0.200}{2}= 0.100 mol}\]

b. `text{pH} = 11.35`

Show Worked Solution

a. $\ce{Ca(OH)2 (s) + 2 HCl (aq) -> CaCl2 (aq) + 2 H2O (l)}$

`text{n(HCl)} = text{c} xx text{V} = 2.00 xx 0.100 = 0.200\ text{mol}`

\[\ce{n(Ca(OH)2) = \frac{n(HCL)}{2} = \frac{0.200}{2}= 0.100 mol}\]

b. $\ce{Ca(OH)2(s) \rightleftharpoons Ca^2+ (aq) + 2 OH– (aq)}$

`[text{Ca}^(2+)] = text{n} / text{V} = 0.100 / 0.100 = 1.00\ text{mol L}^-1`

$\ce{K_{sp}}$	$ \ce{= [Ca^2+][OH– ]^2}$
`5.02 xx 10^(-6)`	`= 1.00 xx [text{OH}^– ]^2`
`[text{OH}^– ]`	`=sqrt{5.02 xx 10^(-6)}`
	`=2.24 xx 10^(−3)\ text{mol L}^(-1)`
`text{pOH }`	`= −log_10(2.24 xx 10^(-3))`
	`= 2.650`

`:.\ text{pH} = 14-2.650 = 11.35`

♦♦♦ Mean mark (b) 20%.

CHEMISTRY, M5 2020 HSC 27

A student makes up a solution of propan-2-amine in water with a concentration of 1.00 mol L ¯¹.

Using structural formulae, complete the equation for the reaction of propan-2-amine with water. (2 marks)
The equilibrium constant for the reaction of propan-2-amine with water is `4.37 xx10^(-4)`.
Calculate the concentration of hydroxide ions in this solution. (3 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

\begin{array} {|l|c|c|c|}
\hline & \ce{C3H7NH2} & \ce{C3H7NH3} & \ce{OH–} \\
\hline \text{Initial} & 1.00 & 0 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 1.00-x & x & x \\
\hline \end{array}

\[ K_b = \ce{\frac{[C3H7NH3+][OH– ]}{[C3H7NH2 ]}} = \frac{x^2}{(1.00-x)} \]

Assume `1.00-x=1.00` because `x` is negligible:

`4.37 xx 10^(−4)`	`= x^2 / 1.00`
`x`	`=sqrt(4.37 xx 10^(−4))`
	`= 0.0209\ text{mol L}^(–1)`

`=> [text{OH}^– ] = 0.0209\ text{mol L}^(–1)`

Show Worked Solution

♦ Mean mark (a) 48%.

\[ K_b = \ce{\frac{[C3H7NH3+][OH– ]}{[C3H7NH2 ]}} = \frac{x^2}{(1.00-x)} \]

Assume `1.00-x=1.00` because `x` is negligible:

`4.37 xx 10^(−4)`	`= x^2 / 1.00`
`x`	`=sqrt(4.37 xx 10^(−4))`
	`= 0.0209\ text{mol L}^(–1)`

`=> [text{OH}^– ] = 0.0209\ text{mol L}^(–1)`

Mean mark (b) 51%.

CHEMISTRY, M6 2020 HSC 14 MC

The equation for the autoionisation of water is shown.

$ \ce{2H2O(l) \rightleftharpoons \ H3O+(aq) + OH-(aq)} $

At 50°C the water ionisation constant, `K_(w)`, is `5.5 xx10^(-14)`.

What is the pH of water at 50°C?

5.50
6.63
6.93
7.00

Show Answers Only

`B`

Show Worked Solution

$\ce{$K_w$ = [H3O+][OH–]}$

Since $\ce{[H3O+] = [OH–]}$:

`[text{H}_3 text{O}^+]^2`	`=5.5 xx 10^(−14)`
`[text{H}_3 text{O}^+]`	`= 2.3 xx 10^(−7}\ text{mol L}^(–1)`

`text{pH = −log}_(10) (2.3 xx 10^(−7) ) = 6.63`

`=> B`

CHEMISTRY, M6 2022 HSC 34

Sodium hypochlorite `\text{NaOCl}` is the active ingredient in pool chlorine. It completely dissolves in water to produce the hypochlorite ion `(\text{OCl}^(-))`, which undergoes hydrolysis according to the following equilibrium.

$ \ce{OCl-(aq) + H2O(l) \rightleftharpoons HOCl(aq) + OH-(aq)} $

The equilibrium constant for this reaction at 25°C is `3.33 xx 10^(-7)`.

For pool chlorine to be effective the pH is maintained by a different buffer at 7.5 and the hypochlorous acid `(\text{HOCl})` concentration should be `1.3 xx 10^(-4)` mol L ¯¹.

Calculate the volume of 2.0 mol L ¯¹ sodium hypochlorite solution that needs to be added to a 1.00 × 10⁴ L pool to meet the required conditions. (4 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`1.3\ text{L}`

Show Worked Solution

`text{pOH}_(eq) = 14.00-7.5 = 6.5`

`[text{OH}^–]_(eq) = 10^-text{pOH} = 10^(-6.5)\ text{mol L}^(-1)`

`text{K}_(eq) `	`=[[text{HOCl}]_(eq) [text{OH}^–]_(eq)] / [text{OCl}^–]_(eq)`
`3.33 xx 10^(−7)`	`= [(1.3 × 10^(−4)) xx (10^(−6.5))] / [[text(OCl)^–]_(eq)]`
`[text(OCl)^–]_(eq)`	`=[(1.3 × 10^(−4)) xx (10^(−6.5))]/(3.33 xx 10^(−7))`
	`= 1.246 xx 10^(−4)\ text{mol L}^(–1)`

\begin{array} {|l|c|c|c|}
\hline & \text{OCl}^– & \text{HOCl} & \text{OH}^– \\
\hline \text{Initial} & x & 0 & – \\
\hline \text{Change} & – 1.3 \times 10^{–4} & +1.3 \times 10^{−4} & – \\
\hline \text{Equilibrium} & x − 1.3 \times 10^{−4} & 1.3 \times 10^{−4} & 10^{−6.5} \\
\hline \end{array}

`x−1.3 xx 10^(−4)`	`= 1.246 xx 10^(−4)`
`x`	`= 2.546 xx 10^(−4)`

`[text{OCl}^–]_i = 2.55 xx 10^(−4)\ text{mol L}^(–1)\ \ text{(3 s.f.)}`

$ \ce{[NaOCl] \times V(NaOC)_{req}}$	$ \ce{= [OCl-](pool) \times V(pool)} $
$ \ce{V(NaCl)_{req}} $	`=(2.55 xx 10^(−4) xx 10^4)/2`
	`=1.3\ text{L (2 s.f.)}`

♦ Mean mark 46%.

CHEMISTRY, M6 2022 HSC 25

The pH of two aqueous solutions was compared.

Explain why the `\text{HCN}(aq)` solution has a higher pH than the `\text{HCl}(aq)` solution. Include a relevant chemical equation for the `\text{HCN}(aq)` solution. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

→ $ \ce{HCl} $ is a strong acid, ie it completely ionises in water to form $ \ce{H+} $ ions.

→ On the other hand, $ \ce{HCN} $ is a weak acid, ie it partially ionises in water to form $ \ce{H+} $ ions.

$ \ce{HCl(aq) -> H+ (aq) + Cl- (aq)} $

$ \ce{HCN(aq) \rightleftharpoons H+ (aq) + CN– (aq)}$

→ As $ \ce{[H+]} $ decreases, pH increases ($ \ce{\text{pH} = – log [H+]} $)

→ Therefore, at the same 0.2M, the $ \ce{HCN} $ solution would have a lower $ \ce{[H+]} $ and thus would have a higher pH than $ \ce{HCl} $.

Show Worked Solution

→ $ \ce{HCl} $ is a strong acid, ie it completely ionises in water to form $ \ce{H+} $ ions.

→ On the other hand, $ \ce{HCN} $ is a weak acid, ie it partially ionises in water to form $ \ce{H+} $ ions.

$ \ce{HCl(aq) -> H+ (aq) + Cl- (aq)} $

$ \ce{HCN(aq) \rightleftharpoons H+ (aq) + CN– (aq)}$

→ As $ \ce{[H+]} $ decreases, pH increases ($ \ce{\text{pH} = – log [H+]} $)

→ Therefore, at the same 0.2M, the $ \ce{HCN} $ solution would have a lower $ \ce{[H+]} $ and thus would have a higher pH than $ \ce{HCl} $.

Mean mark 57%.

CHEMISTRY, M6 2022 HSC 20 MC

Cyanidin is a plant pigment that may be used as a pH indicator. It has four levels of protonation, each with a different colour, represented by these equilibria:

The following graph shows the relative amount of each species present at different pH values.

What colour would the indicator be if added to a 0.75 mol L$^{-1}$ solution of hypoiodous acid, $\text{HIO}\ \left(p K_a=10.64\right)$?

Red
Colourless
Purple
Blue

Show Answers Only

$C$

Show Worked Solution

$\ce{HIO(aq) + H2O(l) \leftrightharpoons IO^– (aq) + H3O^+ (aq)}$

\begin{array} {|l|c|c|c|}
\hline & \text{HIO} & \ \ \text{IO}^– \ \ & \ \ \text{H}_3 \text{O}^+\ \ \\
\hline \text{Initial} & 0.75 & 0 & 0 \\
\hline \text{Change} & -x & +x & +x \\
\hline \text{Equilibrium} & 0.75 − x & x & x \\
\hline \end{array}

$\text{K}_a=\dfrac{\left[ IO ^{-}\right]\left[ H _3 O ^{+}\right]}{[ HIO ]}=\dfrac{x^2}{(0.75-x)}$

$\text{K}_a \ \text{is small} \Rightarrow 0.75-x \approx 0.75$

$\begin{aligned}
K _a & =\dfrac{x^2}{0.75} \\
10^{-10.64} & =\dfrac{x^2}{0.75} \\
x^2 & =10^{-10.64} \times 0.75 \\
x & =\sqrt{10^{-10.64} \times 0.75} \\
& =4.1 \times 10^{-6} \ \text{mol L}^{-1} \\
\therefore\left[ H _3 O ^{+}\right] & =4.1 \times 10^{-6} \ \text{mol L}^{-1}
\end{aligned}$

$\text{pH}=-\log _{10}\left[4.1 \times 10^{-6}\right]=5.38$

$\text{The major species at pH (see graph) = 5.38 is purple.}$

$\Rightarrow C$

♦ Mean mark 46%.

CHEMISTRY, M6 2021 HSC 20 MC

The trimethylammonium ion, $\ce{[({CH_3)_3NH}]^+}$, is a weak acid. The acid dissociation equation is shown.

$\ce{[(CH3)3NH]+($aq$)+H2O($l$)\rightleftharpoons H3O+($aq$)+(CH3)3N($aq$)} \quad K_a = 1.55 \times 10^{-10}$

At 20°C, a saturated solution of trimethylammonium chloride, $\ce{[(CH_3)_3NH]Cl}$, has a pH of 4.46.

What is the $K_{sp}$ of trimethylammonium chloride?

$1.26 \times 10^{-9}$
$7.76$
$60.2$
$5.01 \times 10^{10}$

Show Answers Only

$C$

Show Worked Solution

$\ce{\left[\left(CH3\right)_3 NH \right]^{+}(aq)+ H2O(l) \leftrightharpoons H3O ^{+}(aq)+\left(CH3\right)_3 N(aq)}$

$K_a=\dfrac{\left[\left(\text{CH}_3\right)_3 \text{N}\right]\left[ \text{H}_3 \text{O} ^{+}\right]}{\left[\left( \text{CH} _3\right)_3 \text{NH} \right]^{+}}$

$\text{To calculate \(K_{sp}$ equation:}\)

$\ce{\left[\left(CH _3\right)_3 NH \right] Cl (s) \leftrightharpoons\left[\left( CH _3\right)_3 NH \right]^{+}(aq)+ Cl ^{-}(aq)}$

$K_{sp}=\ce{[(CH3)_3NH)^+] [Cl^-]}$

$\text{pH} = \ce{4.46 \rightarrow \left[H3O^+\right] = 10^{-4.46}}$

$\text{Using stoichiometry;}$

$\ce{[(CH3)_3N)^+]=[H3O^+] = 10^{-4.46}}$

$\text{Using}\ K_{a}:$

$1.55 \times 10^{-10}=\dfrac{\left(10^{-4.46} \times 10^{-4.46}\right)}{\ce{\left(CH3\right)3NH^{+}}}$

$\ce{\left[\left(\left(CH3\right)_3NH \right)^{+}\right]}=\dfrac{\left(10^{-4.46} \times 10^{-4.46}\right)}{1.55 \times 10^{-10}}=7.7565 \ldots \text{mol L}^{-1}$

$\ce{\left[Cl^{-}\right]=\left[\left(\left( CH3\right)_3NH\right)^{+}\right]}=7.7565 \ldots \text{mol L}^{-1}$

$\therefore K_{sp}=\ce{\left[\left(\left(CH3\right)_3NH\right)^{+}\right]\times\left[Cl^{-}\right]=7.7565 \ldots \times 7.7565 \ldots=60.2}$

$\Rightarrow C$

♦♦♦ Mean mark 19%.

\(\ce{pH}\)	\(=-\log_{10}\ce{[H3O+]}\)
\(\ce{[H3O+]}\)	\(=10^{-\ce{pH}}=10^{-1.151}=0.0706\)

\(\ce{pH}\)	\(=-\log_{10}\ce{[H3O+]}\)
\(\ce{[H3O+]}\)	\(=10^{-\ce{pH}}=10^{-1.151}=0.0706\)

\(5.6 \times 10^{-10}\)	\(= \dfrac{\ce{[H3O+]^2}}{0.200} \)
\(\ce{[H3O+]^2}\)	\(=0.200 \times 5.6 \times 10^{-10} \)
\(\ce{[H3O+]}\)	\(= \sqrt{1.12 \times 10^{-10}}\)
	\(=1.06 \times 10^{-5}\ \text{mol L}^{-1} \)

b.	\(K_{a1}\)	\(=\dfrac{\ce{[HS-][H3O+]}}{\ce{[H2S]}} \)
	\(9.6 \times 10^{-8}\)	\(=\dfrac{\ce{[H3O+]^2}}{0.01}\)
	\(\ce{[H3O+]^2}\)	\(=0.01 \times 9.6 \times 10^{-8} \)
	\(\ce{[H3O+]}\)	\(=\sqrt{9.6 \times 10^{-10}}\)
		\(=3.1 \times 10^{-5}\ \text{M} \)

\( \ce{[NaOCl] \times V(NaOC)_{req}}\)	\( \ce{= [OCl-](pool) \times V(pool)} \)
\( \ce{V(NaCl)_{req}} \)	`=(2.55 xx 10^(−4) xx 10^4)/2`
	`=1.3\ text{L (2 s.f.)}`

\(\ce{n(HCl)_{excess}}\)	\(\ce{= n(HCl)_{init}-n(HCl)_{reacted}}\)
	`=0.0500-0.0450`
	`=0.005\ \text{mol}`