CAIE A-Level Chemistry 25. Equilibria Question Bank

Practise equilibrium calculations and explanations involving acids, bases, buffers, conjugate pairs and solubility products.

Covered syllabus points

25.1 Acids and bases 25.2 Partition coefficients

[Maximum number: 1]

Hypophosphorous acid is an inorganic acid.
The conjugate base of hypophosphorous acid is $\mathrm{H}_{2} \mathrm{PO}_{2}^{-}$ .

Give the formula of hypophosphorous acid.

$\mathrm{H}_{3} \mathrm{PO}_{2}$

[Maximum number: 5]

(a)

Define the term partition coefficient, $K_{\mathrm{pc}}$ .

[ 2 ]

(b)

$K_{\mathrm{pc}}$ of benzoic acid between octan-1-ol and water is 79.4.

[ 3 ]

(i)

A solution of 0.400 g of benzoic acid in $25.0 \mathrm{~cm}^{3}$ octan-1-ol is shaken with $125 \mathrm{~cm}^{3}$ of water. Calculate the mass of benzoic acid extracted into the water layer.
mass of benzoic acid extracted =

[ 2 ]

M1 79.4=(0.4-x / 25) /(x / 125)

M2 $\mathrm{x}=0.0237 \mathrm{~g}$ [2] $\min 2 \mathrm{sf}$

(ii)

$\mathrm{K}_{\mathrm{pc}}$ of benzophenone, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COC}_{6} \mathrm{H}_{5}$ , between octan-1-ol and water is different from the value of $K_{\mathrm{pc}}$ of benzoic acid given in (b)(i).

Explain why.

[ 1 ]

[Maximum number: 15]

(a)

Define conjugate acid-base pair.

[ 1 ]

two species that differ by one proton / $\mathrm{H}^{+}$

(b)

Give the formulas of the conjugate acid and the conjugate base of the hydrogen phosphate ion, $\mathrm{HPO}_{4}{ }^{2-}$ .

\begin{aligned} & \text { conjugate acid of } \mathrm{HPO}_{4}^{2-} \\ & \text { conjugate base of } \mathrm{HPO}_{4}^{2-} \end{aligned}

[ 1 ]

$\mathrm{H}_{2} \mathrm{PO}_{4}{ }^{-}$ AND PO4 ${ }^{3-}$

(c)

The $K_{a}$ of propanoic acid, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}$ , is $1.35 \times 10^{-5} \mathrm{moldm}^{-3}$ at 298 K.

Solution C is a solution of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}$ with a pH of 3.60 at 298 K.

[ 7 ]

(i)

Calculate the concentration of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}$ in solution C.

\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . . \mathrm{mol} \mathrm{dm}^{-3}

[2]

[ 2 ]

M1: $\left[\mathrm{H}^{+}\right]=2.51 \times 10^{-4}$

M2: $\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]=4.67 \times 10^{-3}$

(ii)

Calculate the concentration of hydroxide ions in solution C.

\left[\mathrm{OH}^{-}\right]=..............................\ \mathrm{mol\ dm}^{-3}

[ 1 ]

$3.98 \times 10^{-11}$

(iii)

Calculate the concentration of a solution of hydrochloric acid with the same pH as solution C.

\text{concentration}=..............................\ \mathrm{mol\ dm}^{-3}

[ 1 ]

$[\mathrm{HCl}]=2.51 \times 10^{-4}$

(iv)

Table 3.1 shows three possible values of the $K_{a}$ of dimethylpropanoic acid, $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCOOH}$ .

Place a tick in Table 3.1 to show the correct value. Explain your answer.

Table 3.1

explanation

[ 3 ]

(d)

Solution D is made by mixing $100 \mathrm{~cm}^{3}$ of $0.100 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}$ and $100 \mathrm{~cm}^{3}$ of $0.100 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{NaCl}$ .

The pH of solution D is measured as small amounts of $\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})$ are added to it, and when small amounts of NaOH(aq) are added to it.

Solution D only acts as a buffer solution when one of these solutions is added to it.

[ 2 ]

(i)

Complete the sentence and write an equation for the reaction that occurs.

Solution D acts as a buffer when is added to it. equation

[ 1 ]

NaOH AND $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}+\mathrm{OH}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}+\mathrm{H}_{2} \mathrm{O}$

(ii)

Complete the sentence and explain why solution D does not act as a buffer when the other solution is added.

Solution D does not act as a buffer when is added to it. explanation

[ 1 ]

$\mathrm{H}_{2} \mathrm{SO}_{4}$ AND conjugate base of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}$ is not present

(e)

Manganese(II) hydroxide, $\mathrm{Mn}(\mathrm{OH})_{2}$ , is only slightly soluble in water.

The solubility of $\mathrm{Mn}(\mathrm{OH})_{2}$ in water is $3.28 \times 10^{-3} \mathrm{gdm}^{-3}$ at 298 K .

[ 4 ]

(i)

Calculate the concentration of a saturated solution of $\mathrm{Mn}(\mathrm{OH})_{2}$ at 298 K.

\left[\mathrm{Mn}(\mathrm{OH})_{2}\right]=..............................\ \mathrm{mol\ dm}^{-3}

[ 1 ]

$3.69 \times 10^{-5}$

(ii)

Write an expression for the $K_{\mathrm{sp}}$ of $\mathrm{Mn}(\mathrm{OH})_{2}$ . Give the units of $K_{\mathrm{sp}}$ .

K_{s p}=

units =

[ 2 ]

$\text { M1: } K_{\mathrm{sp}}=\left[\mathrm{Mn}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}$

M2: units $=\mathrm{mol}^{3} \mathrm{dm}^{-9}$

(iii)

Use your answers to (d)(i) and (d)(ii) to calculate the value of $K_{sp}$ of $\mathrm{Mn}(\mathrm{OH})_{2}$ at 298 K.

K_{sp}=..............................

[ 1 ]

$2.0 \times 10^{-13}$

[Maximum number: 7]

Iodine is found naturally in compounds in many different oxidation states.

(a)

Iodide ions, $\mathrm{I}^{-}$ , react with acidified $\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})$ to form iodine, $\mathrm{I}_{2}$ , and water. This reaction mixture is shaken with cyclohexane, $\mathrm{C}_{6} \mathrm{H}_{12}$ , to extract the $\mathrm{I}_{2}$ . Cyclohexane is immiscible with water.

[ 4 ]

(i)

$15.0 \mathrm{~cm}^{3}$ of $\mathrm{C}_{6} \mathrm{H}_{12}$ is shaken with $20.0 \mathrm{~cm}^{3}$ of an aqueous solution containing $\mathrm{I}_{2}$ until no further change is seen.
It is found that 0.390 g of $\mathrm{I}_{2}$ is extracted into the $\mathrm{C}_{6} \mathrm{H}_{12}$ .
The partition coefficient of $\mathrm{I}_{2}$ between $\mathrm{C}_{6} \mathrm{H}_{12}$ and water, $K_{\mathrm{pc}}$ , is 93.8.
Calculate the mass of $\mathrm{I}_{2}$ that remains in the aqueous layer.
Show your working.
mass of $I_{2}$ in aqueous layer =

[ 2 ]

(ii)

Suggest how the value of $K_{p c}$ of $\mathrm{I}_{2}$ between hexan-2-one, $\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{COCH}_{3}$ , and water compares to the value given in (a)(ii). Explain your answer.

[ 2 ]

$\quad K_{\mathrm{pc}}$ would be lower

- hexan-2-one is more polar (than cyclohexane)

OR hexan-2-one is polar AND cyclohexane is non-polar

- $\quad \mathrm{I}_{2}$ is (therefore) less soluble in hexan-2-one

All three correct for two marks

(b)

An orange precipitate of $\mathrm{HgI}_{2}$ forms when $\mathrm{Hg}^{2+}$ ions are added to KI(aq). The solubility of $\mathrm{HgI}_{2}$ at $25^{\circ} \mathrm{C}$ is $1.00 \times 10^{-7} \mathrm{gdm}^{-3}$ .

Calculate the solubility product, $K_{\mathrm{sp}}$ , of $\mathrm{HgI}_{2}$ . Include units in your answer.
$\left[M_{\mathrm{r}}: \mathrm{HgI}_{2}, 454.4\right]$
value of $K_{s p}=$
units =

[ 3 ]

\begin{array}{ll} \text { M1: }\left(\left[\mathrm{Hg}^{2+}\right]\right)=1.00 \times 10^{-7} \div 454.4 =2.20 \times 10^{-10}\left(\mathrm{~mol} \mathrm{dm}^{-3}\right) \text { M2: } K_{\mathrm{sp}}=\left[\mathrm{Hg}^{2+}\right]\left[\mathrm{I}^{-}\right]^{2}=4\left[\mathrm{Hg}^{2+}\right]^{3} =\mathbf{4 . 2 6} \times \mathbf{1 0}^{-\mathbf{2 9}} \mathrm{ecf} \text { M3: units }=\mathrm{mol}^{3} \mathrm{dm}^{-9} \mathrm{ecf} \end{array}

All question bank results loaded

Covered syllabus points

25. Equilibria question 1

25. Equilibria question 2

Question (a)

Question (b)

Question (i)

Question (ii)

25. Equilibria question 3

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (iii)

Question (iv)

Question (d)

Question (i)

Question (ii)

Question (e)

Question (i)

Question (ii)

Question (iii)

25. Equilibria question 4

Question (a)

Question (i)

Question (ii)

Question (b)