smc-3671-40-K(eq) and pH

CHEMISTRY, M5 EQ-Bank 23

Butanoic acid is a natural product and a component of human sweat.

Calculate the value of $\ce{$K_{a}$}$ for butanoic acid if a 0.10 mol L^–1 solution has a pH of 2.9 at 298 K. (3 marks)

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$\ce{$K_{a}$ = 1.6 \times 10^{-5}}$

Show Worked Solution

$\ce{C3H7COOH(aq) \rightleftharpoons C3H7COO^-(aq) + H^+(aq)}$ \[\ce{$K_{a}$ = \frac{[C3H7COO^-][H^+]}{[C3H7COOH]}}\]

$\ce{[H^+]}$	$\ce{= 10^{-pH}}$
	$\ce{= 10^{-2.9}}$
	$\ce{= 1.26 \times 10^{-3} mol L^{-1}}$

\begin{array} {|l|c|c|c|}
\hline & \ce{[C3H7COOH]} & \ce{[C3H7COO^-]} & \ce{[H^+]} \\
\hline \text{Initial} & 0.10 & 0 & 0 \\
\hline \text{Change} & -1.26 \times 10^{-3} & +1.26 \times 10^{-3} & +1.26 \times 10^{-3} \\
\hline \text{Equilibrium} & 0.0987 & 1.26 \times 10^{-3} & 1.26 \times 10^{-3} \\
\hline \end{array}

`K_(a)`	`=(1.26 xx10^-3)^2/(0.0987)`
	`=1.6 xx10^(-5)`

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)

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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)

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

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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 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)

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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)

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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)

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\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

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`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`

\(\ce{[H^+]}\)	\(\ce{= 10^{-pH}}\)
	\(\ce{= 10^{-2.9}}\)
	\(\ce{= 1.26 \times 10^{-3} mol L^{-1}}\)