smc-3671-50-Acids and bases

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`

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

`[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%.

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 33

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