Salt hydrolysis — pH of salt solutions (4 cases - SA - SB, SA - WB, WA - SB, WA - WB): Exercises

This page is a self-test ladder. Each problem builds on the parent note Salt hydrolysis (4 cases). Try each problem yourself first, then open the collapsible solution.

Before we start, here is the toolbox — all four master formulas in one place.

What each letter means, from zero:

$K_{w} = 1 0^{- 14}$ is water's self-splitting constant; $p K_{w} = 14$ .
$K_{a}$ measures how willingly a weak acid gives up $H^{+}$ — bigger $K_{a}$ = stronger acid.
$K_{b}$ measures how willingly a weak base grabs $H^{+}$ — bigger $K_{b}$ = stronger base.
$lo g$ here is base-10. $lo g (1 0^{- n}) = - n$ . That is the only log fact we need.

Level 1 — Recognition

(Can you name the case and predict the direction of pH?)

Exercise 1.1

Classify each salt as SA–SB, SA–WB, WA–SB, or WA–WB, and state whether its solution is acidic, basic, or neutral: (a) $KNO_{3}$ (b) $NH_{4} Cl$ (c) $CH_{3} COONa$ (d) $NH_{4} CN$ .

Recall Solution 1.1

Split each salt into its parents (the acid gives the anion, the base gives the cation).

(a) $KNO_{3}$ → $K^{+}$ (from $KOH$ , strong base) + $NO_{3}^{-}$ (from $HNO_{3}$ , strong acid) = SA–SB → neutral, pH = 7.
(b) $NH_{4} Cl$ → $NH_{4}^{+}$ (from $NH_{3}$ , weak base) + $Cl^{-}$ (from $HCl$ , strong acid) = SA–WB → acidic, pH < 7.
(c) $CH_{3} COONa$ → $Na^{+}$ (from $NaOH$ , strong base) + $CH_{3} COO^{-}$ (from $CH_{3} COOH$ , weak acid) = WA–SB → basic, pH > 7.
(d) $NH_{4} CN$ → $NH_{4}^{+}$ (weak base parent) + $CN^{-}$ (from $HCN$ , weak acid) = WA–WB → depends on $K_{a}$ vs $K_{b}$ .

Exercise 1.2

Without any calculator, arrange these solutions (all 0.1 M) in order of increasing pH: $HCl$ 's salt $NaCl$ , $NH_{4} Cl$ , $CH_{3} COONa$ .

Recall Solution 1.2

Acidic sits lowest, neutral in the middle, basic highest.

$NH_{4} Cl$ = acidic → lowest pH
$NaCl$ = neutral → pH = 7
$CH_{3} COONa$ = basic → highest pH

Increasing pH: $NH_{4} Cl < NaCl < CH_{3} COONa$ .

Level 2 — Application

(Plug into the right formula and get a number.)

Exercise 2.1

Find the pH of a 0.1 M $NH_{4} Cl$ solution. Given: $K_{b} (NH_{3}) = 1.8 \times 1 0^{- 5}$ , $K_{w} = 1 0^{- 14}$ .

Recall Solution 2.1

Case: SA–WB (acidic). Use $pH = \frac{1}{2} (p K_{w} - p K_{b} - lo g C)$ .

$p K_{w} = 14$ .
$p K_{b} = - lo g (1.8 \times 1 0^{- 5}) = 4.74$ .
$lo g C = lo g (0.1) = - 1$ .

$pH = \frac{1}{2} (14 - 4.74 - (- 1)) = \frac{1}{2} (14 - 4.74 + 1) = \frac{10.26}{2} = 5.13$ pH ≈ 5.13 (acidic ✓ — as an SA–WB salt must be).

Exercise 2.2

Find the pH of 0.01 M sodium acetate ( $CH_{3} COONa$ ). Given: $K_{a} (CH_{3} COOH) = 1.8 \times 1 0^{- 5}$ , $K_{w} = 1 0^{- 14}$ .

Recall Solution 2.2

Case: WA–SB (basic). Use $pH = \frac{1}{2} (p K_{w} + p K_{a} + lo g C)$ .

$p K_{a} = - lo g (1.8 \times 1 0^{- 5}) = 4.74$ .
$lo g C = lo g (0.01) = - 2$ .

$pH = \frac{1}{2} (14 + 4.74 + (- 2)) = \frac{16.74}{2} = 8.37$ pH ≈ 8.37 (basic ✓).

Exercise 2.3

Find the pH of ammonium acetate ( $CH_{3} COONH_{4}$ ), any concentration. Given: $K_{a} = 1.8 \times 1 0^{- 5}$ , $K_{b} = 1.8 \times 1 0^{- 5}$ .

Recall Solution 2.3

Case: WA–WB. Use $pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$ . Since $K_{a} = K_{b}$ , $p K_{a} = p K_{b} = 4.74$ : $pH = \frac{1}{2} (14 + 4.74 - 4.74) = \frac{14}{2} = 7.0$ pH = 7.0. Note we never needed $C$ — for WA–WB the concentration cancels out entirely.

Level 3 — Analysis

(Reason about how pH changes when inputs change.)

Exercise 3.1

A 0.1 M $NH_{4} Cl$ solution is diluted 100-fold to 0.001 M. Does pH rise or fall, and by how much? (Use $K_{b} = 1.8 \times 1 0^{- 5}$ .)

Recall Solution 3.1

SA–WB formula: $pH = \frac{1}{2} (p K_{w} - p K_{b} - lo g C)$ . Only the $- lo g C$ term changes. Going from $C = 0.1$ to $C = 0.001$ :

$lo g (0.1) = - 1$ , so term $= + 1$ ; contribution to pH $= \frac{1}{2} (+ 1) = + 0.5$ .
$lo g (0.001) = - 3$ , so term $= + 3$ ; contribution $= \frac{1}{2} (+ 3) = + 1.5$ .

Change in pH $= 1.5 - 0.5 = + 1.0$ . pH rises by 1.0 unit (from 5.13 to 6.13). Physical reading: diluting an acidic salt pushes it toward neutral (pH 7), because there is simply less hydrolysing ion around. It approaches 7 but (in this simple model) never overshoots it.

Exercise 3.2

Two 0.1 M basic salts: $CH_{3} COONa$ ( $K_{a} = 1.8 \times 1 0^{- 5}$ ) and $NaCN$ ( $K_{a} (HCN) = 4.9 \times 1 0^{- 10}$ ). Which has the higher pH, and why?

Recall Solution 3.2

WA–SB formula: $pH = \frac{1}{2} (p K_{w} + p K_{a} + lo g C)$ . Same $K_{w}$ , same $C$ — only $p K_{a}$ differs.

Acetate: $p K_{a} = 4.74$ → $pH = \frac{1}{2} (14 + 4.74 - 1) = 8.87$ .
Cyanide: $p K_{a} = - lo g (4.9 \times 1 0^{- 10}) = 9.31$ → $pH = \frac{1}{2} (14 + 9.31 - 1) = 11.16$ .

$NaCN$ has the higher pH (≈ 11.16). Why: a weaker parent acid (smaller $K_{a}$ , bigger $p K_{a}$ ) leaves behind a stronger conjugate base. $HCN$ is far weaker than acetic acid, so $CN^{-}$ grabs $H^{+}$ from water more aggressively → more $OH^{-}$ → higher pH.

Exercise 3.3

Look at the map figure below. Explain in one sentence why the WA–WB line is flat while the other two slope with concentration.

Figure — Salt hydrolysis — pH of salt solutions (4 cases - SA - SB, SA - WB, WA - SB, WA - WB)

Recall Solution 3.3

In SA–WB and WA–SB only one ion hydrolyses, so more salt = more of that ion producing $H^{+}$ or $OH^{-}$ → pH depends on $C$ (sloped lines). In WA–WB, both ions hydrolyse and each dilutes the other's effect proportionally; the $C$ from cation-hydrolysis and the $C$ from anion-hydrolysis cancel algebraically, leaving $pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$ with no $C$ → flat line.

Level 4 — Synthesis

(Combine ideas; work backward; handle the tricky WA–WB.)

Exercise 4.1

A 0.1 M solution of a WA–SB salt has measured pH = 9.0. Find $p K_{a}$ of the parent acid, and hence $K_{a}$ . (Take $K_{w} = 1 0^{- 14}$ .)

Recall Solution 4.1

Work the WA–SB formula backward: $pH = \frac{1}{2} (p K_{w} + p K_{a} + lo g C)$ $9.0 = \frac{1}{2} (14 + p K_{a} + lo g 0.1) = \frac{1}{2} (14 + p K_{a} - 1)$ $18 = 13 + p K_{a} \Rightarrow p K_{a} = 5.0$ Then $K_{a} = 1 0^{- 5.0} = 1.0 \times 1 0^{- 5}$ . Answer: $p K_{a} = 5.0$ , $K_{a} = 1.0 \times 1 0^{- 5}$ .

Exercise 4.2

Predict whether $NH_{4} CN$ is acidic or basic, then compute its pH. Given: $K_{a} (HCN) = 4.9 \times 1 0^{- 10}$ , $K_{b} (NH_{3}) = 1.8 \times 1 0^{- 5}$ , $K_{w} = 1 0^{- 14}$ .

Recall Solution 4.2

Predict: WA–WB. Compare $K_{a}$ vs $K_{b}$ . Here $K_{a} = 4.9 \times 1 0^{- 10}$ (tiny) $<$ $K_{b} = 1.8 \times 1 0^{- 5}$ . The base side wins → basic, pH > 7. Compute with $pH = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$ :

$p K_{a} = - lo g (4.9 \times 1 0^{- 10}) = 9.31$ .
$p K_{b} = - lo g (1.8 \times 1 0^{- 5}) = 4.74$ . $pH = \frac{1}{2} (14 + 9.31 - 4.74) = \frac{1}{2} (18.57) = 9.28$ pH ≈ 9.28 (basic ✓, matching the prediction that the weaker parent acid gives the stronger conjugate base).

Exercise 4.3

For $NH_{4} F$ : $K_{a} (HF) = 6.6 \times 1 0^{- 4}$ , $K_{b} (NH_{3}) = 1.8 \times 1 0^{- 5}$ . Is it acidic or basic? Find pH.

Recall Solution 4.3

WA–WB again. Now $K_{a} = 6.6 \times 1 0^{- 4} > K_{b} = 1.8 \times 1 0^{- 5}$ : the acid side wins → acidic, pH < 7.

$p K_{a} = - lo g (6.6 \times 1 0^{- 4}) = 3.18$ .
$p K_{b} = - lo g (1.8 \times 1 0^{- 5}) = 4.74$ . $pH = \frac{1}{2} (14 + 3.18 - 4.74) = \frac{1}{2} (12.44) = 6.22$ pH ≈ 6.22 (acidic ✓). Note it is only slightly below 7 because neither parent is dramatically stronger than the other.

Level 5 — Mastery

(Multi-step, degenerate cases, and a limit.)

Exercise 5.1

Compute the degree of hydrolysis $h$ for 0.1 M $NH_{4} Cl$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ), and confirm the approximation $h ≪ 1$ used in the parent derivation is justified.

Recall Solution 5.1

$h$ = fraction of the ion that reacted. From the parent derivation, $h = \frac{K _{w}}{K _{b} C}$ . $h = \frac{1 0 ^{- 14}}{( 1.8 \times 1 0 ^{- 5} ) ( 0.1 )} = \frac{1 0 ^{- 14}}{1.8 \times 1 0 ^{- 6}} = 5.56 \times 1 0^{- 9} = 7.45 \times 1 0^{- 5}$ So only about $0.0075%$ of $NH_{4}^{+}$ hydrolyses. Since $h = 7.45 \times 1 0^{- 5} ≪ 1$ , the approximation $C (1 - h) \approx C$ is excellent. ✓ (Cross-check: $[H^{+}] = C h = 0.1 \times 7.45 \times 1 0^{- 5} = 7.45 \times 1 0^{- 6}$ M → pH $= - lo g (7.45 \times 1 0^{- 6}) = 5.13$ , matching Ex 2.1.)

Exercise 5.2 (degenerate input)

What is the pH of a 0.1 M $NaCl$ solution, and what happens to any of our hydrolysis formulas if you naively plugged $NaCl$ into them?

Recall Solution 5.2

$NaCl$ is SA–SB: both ions are spectators, no hydrolysis. pH = 7 exactly (at 25 °C), independent of $C$ . The WA/SB or SA/WB formulas simply do not apply — there is no weak $K_{a}$ or $K_{b}$ to plug in (a strong acid has effectively "infinite" $K_{a}$ , so $p K_{a} \to - \infty$ and the formula breaks). The correct treatment is the trivial one: undisturbed water autoionization gives $[H^{+}] = [OH^{-}] = 1 0^{- 7}$ , pH 7. Lesson: identify the case first; SA–SB is not a limit of the other formulas, it is its own rule.

Exercise 5.3 (limit behaviour)

Using the WA–SB formula, find the pH of $CH_{3} COONa$ at $C = 1 0^{- 4}$ M and comment on the trend as $C \to 0$ . ( $K_{a} = 1.8 \times 1 0^{- 5}$ .)

Recall Solution 5.3

$pH = \frac{1}{2} (14 + 4.74 + lo g (1 0^{- 4})) = \frac{1}{2} (14 + 4.74 - 4) = \frac{1}{2} (14.74) = 7.37$ . pH ≈ 7.37. Compare: at 0.1 M it was 8.87, at 0.01 M it was 8.37, at $1 0^{- 4}$ M just 7.37. Trend: as $C \to 0$ the formula pushes pH toward 7 — dilution empties out the hydrolysing anion so the solution becomes indistinguishable from pure water. (The bare formula would keep sliding below 7 for absurdly small $C$ ; physically it saturates at 7 because water's own $1 0^{- 7}$ dominates — the simple model is only trustworthy down to ~ $1 0^{- 3}$ – $1 0^{- 4}$ M.)

Exercise 5.4 (full synthesis)

A student mislabels $KF$ ( $K_{a} (HF) = 6.6 \times 1 0^{- 4}$ ) as neutral. For a 0.5 M solution, (a) give the correct case, (b) compute the true pH, (c) state by how many units they were wrong.

Recall Solution 5.4

(a) $KF$ → $K^{+}$ (strong base $KOH$ ) + $F^{-}$ (weak acid $HF$ ) = WA–SB → basic, not neutral. (b) $p K_{a} = - lo g (6.6 \times 1 0^{- 4}) = 3.18$ ; $lo g C = lo g (0.5) = - 0.301$ . $pH = \frac{1}{2} (14 + 3.18 + (- 0.301)) = \frac{1}{2} (16.879) = 8.44$ True pH ≈ 8.44. (c) They assumed 7.0; the true value is 8.44, so they were off by about 1.44 pH units — more than a factor of 27 in $[H^{+}]$ .

Recall One-line recap of every answer

1.1 KNO₃ neutral / NH₄Cl acidic / CH₃COONa basic / NH₄CN depends ::: matches parent-splitting rule 2.1 pH of 0.1 M NH₄Cl ::: 5.13 2.2 pH of 0.01 M CH₃COONa ::: 8.37 2.3 pH of CH₃COONH₄ ::: 7.0 3.1 dilution of NH₄Cl 100× ::: pH rises by 1.0 (5.13 → 6.13) 3.2 higher-pH salt ::: NaCN (11.16 vs 8.87) 4.1 back-solve pKa ::: 5.0, Ka = 1.0e-5 4.2 pH of NH₄CN ::: 9.28 (basic) 4.3 pH of NH₄F ::: 6.22 (acidic) 5.1 h for NH₄Cl ::: 7.45e-5 5.3 pH of 1e-4 M acetate ::: 7.37 5.4 pH of 0.5 M KF ::: 8.44