CAIE A-Level Chemistry A2 25.1 Acids & bases

[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: 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: 5]

A student uses thin-layer chromatography (TLC) to analyse a mixture containing different metal cations. The student repeats the experiment using different solvents.

Fig. 6.1 shows the chromatogram obtained by the student using water as a solvent.

Fig. 6.1

(a)

In a third experiment, the pH of the mixture of metal ions is kept constant using a buffer solution.

The student prepares the buffer solution by mixing $20.0 \mathrm{~cm}^{3}$ of $0.150 \mathrm{moldm}^{-3} \mathrm{KOH}(\mathrm{aq})$ and $50.0 \mathrm{~cm}^{3}$ of $0.100 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4} \mathrm{~K}(\mathrm{aq})$ .
$\mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4} \mathrm{~K}$ is a weak carboxylic acid that has $\mathrm{pK}_{\mathrm{a}}=5.40$ .

Fig. 6.2

[ 5 ]

(i)

Complete the equation for the reaction of $\mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4} \mathrm{~K}(\mathrm{aq})$ with KOH(aq).
$\mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4} \mathrm{~K}+$

[ 1 ]

$\mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4} \mathrm{~K}+\mathrm{KOH} \rightarrow \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4} \mathrm{~K}_{2}+\mathrm{H}_{2} \mathrm{O} \quad$ OR $\mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4} \mathrm{~K}+\mathrm{OH}^{-} \rightarrow \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4} \mathrm{~K}^{-}+\mathrm{H}_{2} \mathrm{O}$

(ii)

Calculate the pH of the buffer solution. Show all your working.

\mathrm{pH}=

[ 4 ]

M1 initially: moles of $\mathrm{KOH}=0.150 \times 20.0 \div 1000 \quad=3.0 \times 10^{-3}$

AND moles of acid HA $=0.100 \times 50.0 \div 1000=5.0 \times 10^{-3}$

M2 at equilibrium: moles of salt KA $\quad=3.0 \times 10^{-3}$ ecf

AND moles of acid $\mathrm{HA}=5.0 \times 10^{-3}$ - moles of NaOH $=2.0 \times 10^{-3}$ ecf

$\begin{aligned} \text { M3 } K_{a}=10^{-5.40}=3.98 \times 10^{-6} 3.98 \times 10^{-6}=\left[\mathrm{H}^{+}\right]\left[3.0 \times 10^{-3}\right] \div\left[2.0 \times 10^{-3}\right] {\left[\mathrm{H}^{+}\right]=2.65 \times 10^{-6} \text { eff } } \end{aligned}$

M4 $\mathrm{pH}=-\log \left(2.65 \times 10^{-6}\right)=\mathbf{5 . 5 8} / \mathbf{5 . 6}$ ecf on a calculated and identified $\left[\mathrm{H}^{+}\right]$

ALLOW alternative approach using Henderson-Hasselbalch equation for M3 and M4

[Maximum number: 4]

A student uses thin-layer chromatography (TLC) to analyse a mixture containing different metal cations. The student repeats the experiment using different solvents.

Fig. 6.1 shows the chromatogram obtained by the student using water as a solvent.

Fig. 6.1

In a third experiment, the pH of the mixture of metal ions is kept constant using a buffer solution.

The student prepares the buffer solution by mixing $20.0 \mathrm{~cm}^{3}$ of $0.150 \mathrm{moldm}^{-3} \mathrm{KOH}(\mathrm{aq})$ and $50.0 \mathrm{~cm}^{3}$ of $0.100 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4} \mathrm{~K}(\mathrm{aq})$ .
$\mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4} \mathrm{~K}$ is a weak carboxylic acid that has $\mathrm{pK}_{\mathrm{a}}=5.40$ .

Fig. 6.2

Calculate the pH of the buffer solution. Show all your working.

\mathrm{pH}=

M1 initially: moles of $\mathrm{KOH}=0.150 \times 20.0 \div 1000 \quad=3.0 \times 10^{-3}$

AND moles of acid HA $=0.100 \times 50.0 \div 1000=5.0 \times 10^{-3}$

M2 at equilibrium: moles of salt KA $\quad=3.0 \times 10^{-3}$ ecf

AND moles of acid $\mathrm{HA}=5.0 \times 10^{-3}$ - moles of NaOH $=2.0 \times 10^{-3}$ ecf

$\begin{aligned} \text { M3 } K_{a}=10^{-5.40}=3.98 \times 10^{-6} 3.98 \times 10^{-6}=\left[\mathrm{H}^{+}\right]\left[3.0 \times 10^{-3}\right] \div\left[2.0 \times 10^{-3}\right] {\left[\mathrm{H}^{+}\right]=2.65 \times 10^{-6} \text { eff } } \end{aligned}$

M4 $\mathrm{pH}=-\log \left(2.65 \times 10^{-6}\right)=\mathbf{5 . 5 8} / \mathbf{5 . 6}$ ecf on a calculated and identified $\left[\mathrm{H}^{+}\right]$

ALLOW alternative approach using Henderson-Hasselbalch equation for M3 and M4

[Maximum number: 2]

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 .

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 =

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

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

[Maximum number: 3]

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

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 =

\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}

[Maximum number: 2]

Calcium iodate $(\mathrm{V}), \mathrm{Ca}\left(\mathrm{IO}_{3}\right)_{2}$ , is sparingly soluble in water.

The concentration of its saturated solution is $5.6 \times 10^{-3} \mathrm{~mol} \mathrm{dm}^{-3}$ at 298 K .

When a few $\mathrm{cm}^{3}$ of concentrated $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})$ is added to a saturated solution of $\mathrm{Ca}\left(\mathrm{IO}_{3}\right)_{2}$ a white precipitate forms.

Identify the white precipitate and give an explanation for this observation.

M1 $\mathrm{Ca}\left(\mathrm{IO}_{3}\right)_{2}$ AND as solubility of $\mathrm{Ca}\left(\mathrm{IO}_{3}\right)_{2}$ decreases

M2 due to common ion effect

CAIE A-Level Chemistry A2 25.1 Acids and Bases Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question 2

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)

Question 3

Question (a)

Question (i)

Question (ii)

Question 4

Question 5

Question 6

Question 7