pH of Acidic and Alkaline Solutions

Lessons

Senior 6 · Chemistry

pH of Acidic and Alkaline Solutions

5 lessons

Lesson 4 of 5

Salt Hydrolysis and the pH of Salt Solutions

25 min

Salts that are not neutral

A salt is what forms when an acid neutralises a base, so you might expect every salt solution to have pH 7. Test some 0.10 mol dm⁻³ solutions with a pH meter at 25 °C and you find otherwise:

Salt solution (0.10 mol dm⁻³)	pH	Made from
sodium chloride, NaCl	7.0	strong acid + strong base
ammonium chloride, $\mathrm{NH_4Cl}$	5.1	strong acid + weak base
sodium ethanoate, $\mathrm{CH_3COONa}$	8.9	weak acid + strong base
sodium carbonate, $\mathrm{Na_2CO_3}$	11.7	weak acid + strong base
aluminium chloride, $\mathrm{AlCl_3}$	about 3	strong acid + weak base

Table 4.1: The pH of some salt solutions, calculated from the constants in Lesson 3.

Ammonium fertilisers make soils more acidic over time, sodium carbonate (washing soda) is alkaline enough to attack grease, and aluminium sulfate, added to water at treatment works, makes the dirt in it clump together. All three effects come from the reaction of an ion with water: salt hydrolysis.

Definition

Salt hydrolysis: the reaction of the cation or anion of a salt with water, producing $\mathrm{H_3O^{+}}$ or $\mathrm{OH^{-}}$ , so that the solution is not neutral. Example: $\mathrm{CH_3COO^{-} + H_2O \rightleftharpoons CH_3COOH + OH^{-}}$ .

Hydrolysis constant, $K_{h}$ : the equilibrium constant for a hydrolysis reaction. For the anion of a weak acid HA, $K_h = K_b(\mathrm{A^{-}}) = \frac{K_w}{K_a}$ ; for the cation of a weak base B, $K_h = K_a(\mathrm{BH^{+}}) = \frac{K_w}{K_b}$ .

Spectator ion: an ion that is the conjugate of a strong acid or base (e.g. $\mathrm{Na^{+}}$ , $\mathrm{K^{+}}$ , $\mathrm{Cl^{-}}$ , $\mathrm{NO_3^{-}}$ , $\mathrm{SO_4^{2-}}$ ); it is too weak an acid or base to react with water.

Which ions hydrolyse?

Use the rule from Lesson 3: the weaker an acid, the stronger its conjugate base.

Strong acid + strong base (NaCl, $\mathrm{KNO_3}$ , $\mathrm{Na_2SO_4}$ ): neither ion reacts with water. The solution is neutral, pH 7 at 25 °C.
Weak acid + strong base ( $\mathrm{CH_3COONa}$ , $\mathrm{Na_2CO_3}$ , KCN): the anion is a base and takes a proton from water. The solution is alkaline:
$\mathrm{CO_3^{2-}(aq) + H_2O(l) \rightleftharpoons HCO_3^{-}(aq) + OH^{-}(aq)}$
Strong acid + weak base ( $\mathrm{NH_4Cl}$ , $\mathrm{(NH_4)_2SO_4}$ ): the cation is an acid and gives a proton to water. The solution is acidic:
$\mathrm{NH_4^{+}(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^{+}(aq)}$
Weak acid + weak base ( $\mathrm{CH_3COONH_4}$ ): both ions react. The pH depends on which is stronger; for ammonium ethanoate, p $K_a(\mathrm{NH_4^{+}}) = 9.25$ and p $K_b(\mathrm{CH_3COO^{-}}) = 9.24$ are almost equal, so the solution is close to pH 7.

Small, highly charged metal ions

Some metal cations are acids too. In water, $\mathrm{Al^{3+}}$ is surrounded by six water molecules, $\mathrm{[Al(H_2O)_6]^{3+}}$ . The small, highly charged ion pulls electron density towards itself, weakening the O–H bonds of its water ligands, so one of them can lose a proton to the solvent (Figure 4.1). Its $K_{a}$ is about $10^{-5}\,\mathrm{mol\,dm^{-3}}$ , close to that of ethanoic acid. $\mathrm{[Fe(H_2O)_6]^{3+}}$ behaves in the same way, while $\mathrm{Na^{+}}$ and $\mathrm{K^{+}}$ , large and singly charged, do not.

Figure 4.1: Why the hexaaquaaluminium ion is acidic.

This is why aluminium sulfate makes water slightly acidic, and why, as the hydrolysis continues, insoluble aluminium hydroxide forms and sweeps fine dirt out of the water. In very acidic soils (below about pH 5) the reverse happens: aluminium dissolves as $\mathrm{Al^{3+}}$ ions that damage plant roots, one reason lime is applied.

Calculating the pH of a salt solution

Treat the hydrolysing ion as a weak base (or weak acid) of concentration $c$ , with $K_b = \frac{K_w}{K_a}$ (or $K_a = \frac{K_w}{K_b}$ ), and use the method of Lesson 3:

$\text{salt of a weak acid: } [\mathrm{OH^{-}}] = \sqrt{\frac{K_w}{K_a}\,c} \qquad \text{salt of a weak base: } [\mathrm{H_3O^{+}}] = \sqrt{\frac{K_w}{K_b}\,c}$

Taking logarithms gives two forms that are useful for checking at 25 °C:

$\mathrm{pH} = 7 + \tfrac{1}{2}\mathrm{p}K_a + \tfrac{1}{2}\log_{10}c \qquad \mathrm{pH} = 7 - \tfrac{1}{2}\mathrm{p}K_b - \tfrac{1}{2}\log_{10}c$

Worked example

Problem. Calculate the pH of 0.100 mol dm⁻³ sodium ethanoate at 25 °C ( $K_a(\mathrm{CH_3COOH}) = 1.74 \times 10^{-5}\,\mathrm{mol\,dm^{-3}}$ ).

Solution. $\mathrm{Na^{+}}$ is a spectator. The ethanoate ion hydrolyses: $\mathrm{CH_3COO^{-} + H_2O \rightleftharpoons CH_3COOH + OH^{-}}$ .

$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.75 \times 10^{-10}\,\mathrm{mol\,dm^{-3}}$

$[\mathrm{OH^{-}}] = \sqrt{5.75 \times 10^{-10} \times 0.100} = 7.58 \times 10^{-6}\,\mathrm{mol\,dm^{-3}} \qquad \mathrm{pOH} = 5.12$

Check with the formula: $\mathrm{pH} = 7 + \tfrac{1}{2}(4.76) + \tfrac{1}{2}\log_{10}(0.100) = 7 + 2.38 - 0.50 = 8.88$ .

Answer: pH = 14.00 − 5.12 = 8.88. Only 0.008 % of the ethanoate ions are hydrolysed, but that is enough to make the solution clearly alkaline.

Worked example

Problem. (a) Calculate the pH of 0.0500 mol dm⁻³ ammonium nitrate ( $K_b(\mathrm{NH_3}) = 1.8 \times 10^{-5}$ ). (b) What mass of ammonium chloride ( $M = 53.49$ ) must be dissolved to make 250 cm³ of solution of pH 5.00?

Solution.

(a) $K_a(\mathrm{NH_4^{+}}) = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$ .

$[\mathrm{H_3O^{+}}] = \sqrt{5.6 \times 10^{-10} \times 0.0500} = 5.3 \times 10^{-6}\,\mathrm{mol\,dm^{-3}} \qquad \mathrm{pH} = 5.28$

(b) At pH 5.00, $[\mathrm{H_3O^{+}}] = 1.0 \times 10^{-5}$ . From $[\mathrm{H_3O^{+}}]^2 = K_a c$ :

$c = \frac{(1.0 \times 10^{-5})^2}{5.6 \times 10^{-10}} = 0.18\,\mathrm{mol\,dm^{-3}}$

Mass $= 0.18 \times 0.250 \times 53.49 = 2.4\,\mathrm{g}$ .

Answer: (a) pH 5.28; (b) about 2.4 g. This is why ammonium fertilisers, used year after year, can acidify a soil.

Warning

Common mistakes

Assuming every salt is neutral. Ask first: is the parent acid weak, is the parent base weak?
Using $K_{a}$ of the acid directly for its anion. The anion is a base: use $K_b = \frac{K_w}{K_a}$ .
Writing hydrolysis of a spectator ion, e.g. $\mathrm{Cl^{-} + H_2O \rightarrow HCl + OH^{-}}$ . HCl is a strong acid, so $\mathrm{Cl^{-}}$ has no tendency to take a proton from water.
Confusing a salt of a weak acid with a buffer. Sodium ethanoate on its own is not a buffer; mixed with ethanoic acid it is (Lesson 5).

Tip

"The strong parent wins." A salt's solution takes the character of its stronger parent: strong acid parent → acidic, strong base parent → alkaline, both strong → neutral. Then calculate with the ion that comes from the weak parent, because that is the one that reacts with water.

Making Sense Chem (Indigo Education Group): Acids and Bases, Salt Hydrolysis. The video begins with an experiment testing salt solutions with indicator. Watch for which ions react with water, and why the solutions are acidic, alkaline or neutral.

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Practice

Predict whether each 0.1 mol dm⁻³ solution is acidic, alkaline or neutral, and write an equation for any hydrolysis: (a) $\mathrm{KNO_3}$ ; (b) $\mathrm{K_2CO_3}$ ; (c) $\mathrm{(NH_4)_2SO_4}$ ; (d) $\mathrm{FeCl_3}$ .
Calculate the pH of 0.200 mol dm⁻³ sodium benzoate ( $K_a(\mathrm{C_6H_5COOH}) = 6.3 \times 10^{-5}$ ).
Calculate the pH of 0.0100 mol dm⁻³ potassium cyanide ( $K_a(\mathrm{HCN}) = 6.2 \times 10^{-10}$ ). Is the approximation justified?
Show that the pH of 0.10 mol dm⁻³ sodium carbonate is about 11.7, given that $K_a(\mathrm{HCO_3^{-}}) = 4.7 \times 10^{-11}$ .
Explain why a solution of aluminium chloride is acidic, while a solution of sodium chloride is neutral.
Use the formula $\mathrm{pH} = 7 + \tfrac{1}{2}\mathrm{p}K_a + \tfrac{1}{2}\log_{10}c$ to predict how the pH of a sodium ethanoate solution changes when it is diluted tenfold.
A farmer has a choice of two nitrogen fertilisers for an already acidic field: ammonium sulfate or calcium nitrate. Evaluate which is the better choice, using salt hydrolysis.

Show answersHide answers

(a) neutral (both parents strong); (b) alkaline, $\mathrm{CO_3^{2-} + H_2O \rightleftharpoons HCO_3^{-} + OH^{-}}$ ; (c) acidic, $\mathrm{NH_4^{+} + H_2O \rightleftharpoons NH_3 + H_3O^{+}}$ ; (d) acidic, $\mathrm{[Fe(H_2O)_6]^{3+} + H_2O \rightleftharpoons [Fe(H_2O)_5(OH)]^{2+} + H_3O^{+}}$ .
$K_b = 1.6 \times 10^{-10}$ ; $[\mathrm{OH^{-}}] = \sqrt{1.6 \times 10^{-10} \times 0.200} = 5.6 \times 10^{-6}$ ; pOH = 5.25, pH = 8.75.
$K_b(\mathrm{CN^{-}}) = 1.6 \times 10^{-5}$ ; $[\mathrm{OH^{-}}] = \sqrt{1.6 \times 10^{-5} \times 0.0100} = 4.0 \times 10^{-4}$ ; pH = 10.60. $\alpha = 4\,\%$ , just under 5 %, so the approximation is acceptable.
$K_b(\mathrm{CO_3^{2-}}) = \frac{1.0 \times 10^{-14}}{4.7 \times 10^{-11}} = 2.1 \times 10^{-4}$ ; $[\mathrm{OH^{-}}] = \sqrt{2.1 \times 10^{-4} \times 0.10} = 4.6 \times 10^{-3}$ ; pOH = 2.34, pH = 11.66 ≈ 11.7.
$\mathrm{Al^{3+}}$ is small and highly charged; in $\mathrm{[Al(H_2O)_6]^{3+}}$ it polarises the O–H bonds of the water ligands, which then release $\mathrm{H^{+}}$ to the solvent. $\mathrm{Na^{+}}$ is larger with only one charge, and $\mathrm{Cl^{-}}$ is the conjugate base of a strong acid, so neither reacts with water.
$\log_{10}c$ falls by 1, so the pH falls by $\tfrac{1}{2}$ : from 8.88 to 8.38. The solution stays alkaline but becomes less so.
Ammonium sulfate is the salt of a strong acid and a weak base: $\mathrm{NH_4^{+}}$ hydrolyses to give $\mathrm{H_3O^{+}}$ (and, in soil, its conversion to nitrate releases more acid), so it would make the field more acidic. Calcium nitrate comes from a strong acid and a strong base and does not hydrolyse. Calcium nitrate is the better choice for an acidic field, unless lime is also applied.

Sources and credits

Dissociation constants: values as tabulated in the CRC Handbook of Chemistry and Physics; the pH values in Table 4.1 were calculated by Ganzaa.
Image: Floculação.jpg by zardeto, via Wikimedia Commons, public domain.
Video: Acids and Bases, Salt Hydrolysis, A Level H2 Chem, Making Sense Chem (Indigo Education Group), embedded from YouTube.
Written by Ganzaa: the text, diagram, tables, worked examples and Practice items.

1Salts that are not neutral

2Which ions hydrolyse?

Small, highly charged metal ions

3Calculating the pH of a salt solution

Salts that are not neutral

Which ions hydrolyse?

Calculating the pH of a salt solution