pH, pOH, pKa, pKb scales: Exercises — Chemistry deep dive

This page is a self-test. Each problem sits above a collapsible solution — read the problem, try it on paper, then expand. We climb five rungs: L1 Recognition → L2 Application → L3 Analysis → L4 Synthesis → L5 Mastery. Every tool used here was built in the parent note the pH/pOH/pKa/pKb topic and its prerequisites; when a symbol appears, we re-anchor it in one line so you never meet it cold.

The number line above is our map for the whole page: acidic to the left, neutral at 7, basic to the right. Keep glancing at it — every answer should land somewhere on it and feel right.

L1 — Recognition

Here you only need to recognise the correct definition and plug in. No multi-step reasoning yet.

Exercise 1.1

A solution has $[H^{+}] = 1 0^{- 5} M$ . State its pH.

Recall Solution

What we do: apply the definition $pH = - lo g_{10} [H^{+}]$ . Why: pH is defined as the negative power of ten of the hydrogen-ion concentration — nothing more. $pH = - lo g_{10} (1 0^{- 5}) = - (- 5) = 5$ Read the map: pH 5 sits left of 7 → acidic. Correct, since $1 0^{- 5} > 1 0^{- 7}$ (more $H^{+}$ than pure water).

Exercise 1.2

A solution has $pOH = 9$ . Find its pH.

Recall Solution

What we do: use $pH + pOH = 14$ . Why this tool: the water autoionisation constant $K_{w} = [H^{+}] [OH^{-}] = 1 0^{- 14}$ forces the two p-values to add to 14 (see 2.6.5-Ionic-product-of-water-Kw). $pH = 14 - pOH = 14 - 9 = 5$ pH 5 → acidic. A solution with high pOH (little $OH^{-}$ ) is indeed acidic. ✓

Exercise 1.3

A weak acid has $K_{a} = 1 0^{- 4}$ . State its $pK_{a}$ .

Recall Solution

What we do: $pK_{a} = - lo g_{10} (K_{a})$ . $pK_{a} = - lo g_{10} (1 0^{- 4}) = 4$ A larger $pK_{a}$ means a weaker acid; $pK_{a} = 4$ is a moderately weak acid.

L2 — Application

Now you run the definition backwards (exponentials) or chain two steps.

Exercise 2.1

A solution has $pH = 3.5$ . Find $[H^{+}]$ .

Recall Solution

What we do: invert the p-operator. Since $pH = - lo g_{10} [H^{+}]$ , exponentiate: $[H^{+}] = 1 0^{- pH} = 1 0^{- 3.5}$ Why the power of ten: $1 0^{x}$ undoes $lo g_{10}$ — they are inverse operations, like squaring undoes square-root. Splitting $1 0^{- 3.5} = 1 0^{- 4} \times 1 0^{0.5} = 1 0^{- 4} \times 3.162$ : $[H^{+}] = 3.16 \times 1 0^{- 4} M$

Exercise 2.2

$[OH^{-}] = 2.0 \times 1 0^{- 3} M$ . Find the pH.

Recall Solution

Step 1 — get pOH. $pOH = - lo g_{10} (2.0 \times 1 0^{- 3})$ . Use the product rule $lo g (ab) = lo g a + lo g b$ (this turns an awkward product into a sum): $pOH = - (lo g 2.0 + lo g 1 0^{- 3}) = - (0.301 - 3) = 2.70$ Step 2 — cross to pH with $pH + pOH = 14$ : $pH = 14 - 2.70 = 11.30$ Sanity check on the map: pH 11.3 is well to the right → basic. Correct, because $[OH^{-}] = 2 \times 1 0^{- 3}$ is far above the neutral $1 0^{- 7}$ .

Exercise 2.3

Acetic acid has $pK_{a} = 4.76$ . Find $pK_{b}$ of its conjugate base, acetate.

Recall Solution

What we do: conjugate pairs obey $pK_{a} + pK_{b} = 14$ , because $K_{a} \cdot K_{b} = K_{w}$ . $pK_{b} = 14 - 4.76 = 9.24$ $pK_{b} = 9.24 > 7$ → acetate is a weak base. A weak acid always begets a weak conjugate base.

L3 — Analysis

Here you compare, rank, and interpret — not just compute one number.

Exercise 3.1

Acid A has $pK_{a} = 3.2$ ; Acid B has $pK_{a} = 5.8$ . Which is stronger, and by what factor in $K_{a}$ ?

Recall Solution

Which is stronger: lower $pK_{a}$ ⇒ larger $K_{a}$ ⇒ more dissociation ⇒ Acid A is stronger. By how much: the gap is $Δ pK_{a} = 5.8 - 3.2 = 2.6$ . Each unit of $pK_{a}$ is a factor of 10 in $K_{a}$ , so: $\frac{K _{a} ( A )}{K _{a} ( B )} = 1 0^{2.6} \approx 398 \approx 400$ Why the ratio is $1 0^{Δ}$ : subtracting logs = dividing the originals, $1 0^{l o g a - l o g b} = a / b$ . Acid A is roughly 400× stronger.

Exercise 3.2

Two solutions: one at pH 4, one at pH 6. How many times more $H^{+}$ is in the first?

Recall Solution

$[H^{+}] = 1 0^{- pH}$ , so the ratio is: $\frac{1 0 ^{- 4}}{1 0 ^{- 6}} = 1 0^{- 4 - (- 6)} = 1 0^{2} = 100$ A 2-unit drop in pH = 100× more acidic. This is why pH is logarithmic: small number changes hide huge concentration changes.

Exercise 3.3

A base has $pK_{b} = 4.75$ (ammonia). Is its conjugate acid (ammonium) stronger or weaker than acetic acid ( $pK_{a} = 4.76$ )?

Recall Solution

Step 1 — get ammonium's $pK_{a}$ using $pK_{a} + pK_{b} = 14$ : $pK_{a} (NH_{4}^{+}) = 14 - 4.75 = 9.25$ Step 2 — compare. Acetic acid $pK_{a} = 4.76$ is much lower than $9.25$ , so acetic acid is the stronger acid (ammonium is very weak). Interpretation: a fairly strong base (ammonia) has a very weak conjugate acid — exactly what $pK_{a} + pK_{b} = 14$ predicts.

L4 — Synthesis

Now you combine the p-scales with equilibrium ideas from 2.6.7-Buffer-solutions and 2.6.8-Henderson-Hasselbalch-equation.

Exercise 4.1

A buffer contains equal moles of acetic acid and acetate ( $pK_{a} = 4.76$ ). What is the pH?

Recall Solution

Tool: the Henderson–Hasselbalch equation from 2.6.8-Henderson-Hasselbalch-equation: $pH = pK_{a} + lo g_{10} \frac{[ A ^{-} ]}{[ HA ]}$ Why this tool: it links the fixed acid constant $pK_{a}$ to the actual pH via the ratio of base to acid — perfect for a buffer. Equal moles ⇒ ratio $= 1$ ⇒ $lo g_{10} (1) = 0$ : $pH = 4.76 + 0 = 4.76$ This is the buffer midpoint: at $pH = pK_{a}$ exactly half the acid is dissociated (see figure below).

What this figure plots and why it matters: the horizontal axis is the solution's pH; the two curves show what fraction of the total acid exists as $HA$ (amber, undissociated) versus $A^{-}$ (cyan, dissociated). As pH rises, the amber curve falls and the cyan curve rises — they cross at exactly $pH = pK_{a} = 4.76$ , where each equals $0.5$ . That crossing point is the buffer midpoint of Exercise 4.1: equal amounts of acid and conjugate base. Notice the curves are steepest and flattest-in-slope near the crossing, which is why a buffer resists pH change best there — a small addition of acid or base only nudges the ratio slightly.

Exercise 4.2

You want a buffer at pH 5.00 from the same acetic-acid system. What ratio $[A^{-}] / [HA]$ is needed?

Recall Solution

Rearrange Henderson–Hasselbalch for the ratio: $lo g_{10} \frac{[ A ^{-} ]}{[ HA ]} = pH - pK_{a} = 5.00 - 4.76 = 0.24$ Undo the log: $\frac{[ A ^{-} ]}{[ HA ]} = 1 0^{0.24} \approx 1.74$ Interpretation: to push pH slightly above $pK_{a}$ , you need slightly more conjugate base than acid. Makes sense — more base means less acidic.

Exercise 4.3

Pure water at $2 5^{\circ} C$ has $pH = 7$ . A student cools water so that $K_{w} = 1 0^{- 15}$ . What is the pH of neutral water now?

Recall Solution

Re-anchor $p K_{w}$ : apply the p-operator to the water constant, $p K_{w} = - lo g_{10} (K_{w})$ . At $2 5^{\circ} C$ , $K_{w} = 1 0^{- 14}$ so $p K_{w} = 14$ ; here $K_{w} = 1 0^{- 15}$ so $p K_{w} = 15$ . Key idea: "neutral" means $[H^{+}] = [OH^{-}]$ , not "pH = 7". The number 7 came from $K_{w} = 1 0^{- 14}$ at $2 5^{\circ} C$ . Set $[H^{+}] = [OH^{-}] = x$ : $K_{w} = x \cdot x = x^{2} = 1 0^{- 15} \Rightarrow x = 1 0^{- 7.5}$ $pH = - lo g_{10} (1 0^{- 7.5}) = 7.5$ Neutral water now has pH 7.5 — still neutral because $[H^{+}] = [OH^{-}]$ , even though the number isn't 7. Note $pH = \frac{1}{2} p K_{w} = \frac{15}{2} = 7.5$ .

L5 — Mastery

Full multi-stage problems mixing weak-acid equilibria (2.6.12-pH-calculations-for-weak-acids-bases) with the p-scales.

Exercise 5.1

Compute the pH of a $0.10 M$ solution of a weak acid HA with $pK_{a} = 4.76$ .

Recall Solution

Step 1 — get $K_{a}$ : $K_{a} = 1 0^{- pK_{a}} = 1 0^{- 4.76} = 1.74 \times 1 0^{- 5}$ . Step 2 — set up the equilibrium $HA ⇌ H^{+} + A^{-}$ . Let $x = [H^{+}]$ formed: $K_{a} = \frac{x \cdot x}{0.10 - x} \approx \frac{x ^{2}}{0.10} (since x ≪ 0.10)$ Why the approximation: a weak acid dissociates little, so $x$ barely dents the starting $0.10 M$ — dropping it keeps the algebra a simple square root. Step 3 — solve for $x$ : $x = K_{a} \times 0.10 = 1.74 \times 1 0^{- 5} \times 0.10 = 1.74 \times 1 0^{- 6} = 1.32 \times 1 0^{- 3} M$ Step 4 — take pH: $pH = - lo g_{10} (1.32 \times 1 0^{- 3}) = 2.88$ Check the approximation: $x /0.10 = 1.3% < 5%$ ✓, so dropping $x$ was valid.

Exercise 5.2

A $0.050 M$ solution of a weak base B has $pK_{b} = 4.75$ . Find the pH.

Recall Solution

Step 1 — $K_{b}$ : $K_{b} = 1 0^{- 4.75} = 1.78 \times 1 0^{- 5}$ . Step 2 — equilibrium $B + H_{2} O ⇌ BH^{+} + OH^{-}$ , let $x = [OH^{-}]$ : $K_{b} \approx \frac{x ^{2}}{0.050} \Rightarrow x = 1.78 \times 1 0^{- 5} \times 0.050 = 9.43 \times 1 0^{- 4} M$ Step 3 — pOH then pH: $pOH = - lo g_{10} (9.43 \times 1 0^{- 4}) = 3.03, pH = 14 - 3.03 = 10.97$ Read the map: pH ≈ 11 → basic, as a base solution must be. ✓

Exercise 5.3

A solution is made $1.0 \times 1 0^{- 8} M$ in HCl. Show that its pH is not 8, and find the true pH.

Recall Solution

Why naïve pH 8 is impossible: pH 8 would be basic, yet we added acid! The mistake is ignoring the $H^{+}$ that water itself supplies ( $1 0^{- 7} M$ ), which now dominates the tiny added amount. Charge/mass balance gives total $[H^{+}]$ satisfying (with $K_{w} = 1 0^{- 14}$ ): $[H^{+}] = C_{acid} + \frac{K _{w}}{[ H ^{+} ]} \Rightarrow [H^{+}]^{2} - C_{acid} [H^{+}] - K_{w} = 0$ with $C_{acid} = 1.0 \times 1 0^{- 8}$ . Solve the quadratic (positive root): $[H^{+}] = \frac{C _{acid} + C _{acid}^{2} + 4 K _{w}}{2} = \frac{1 0 ^{- 8} + 1 0 ^{- 16} + 4 \times 1 0 ^{- 14}}{2} = 1.05 \times 1 0^{- 7} M$ $pH = - lo g_{10} (1.05 \times 1 0^{- 7}) = 6.98$ Result: pH ≈ 6.98, just barely acidic — exactly right for an ultra-dilute strong acid. "Strong" means fully dissociated, not "always low pH" (see 2.6.11-pH-calculationsfor-strong-acids-bases).

Recall Self-test checklist (expand only after attempting all)

p-operator ::: $p X = - lo g_{10} (X)$ — applies to any quantity, e.g. $p K_{w} = - lo g_{10} (K_{w})$ Forward pH ::: $pH = - lo g_{10} [H^{+}]$ Reverse pH ::: $[H^{+}] = 1 0^{- pH}$ Cross to pOH ::: $pH + pOH = 14$ (25 °C) Conjugate link ::: $pK_{a} + pK_{b} = 14$ (25 °C) Buffer pH ::: $pH = pK_{a} + lo g_{10} \frac{[ A ^{-} ]}{[ HA ]}$ Neutral condition ::: $[H^{+}] = [OH^{-}]$ , i.e. $pH = \frac{1}{2} p K_{w}$ (only 7 at 25 °C, where $p K_{w} = 14$ ) Can pH leave 0–14? ::: yes — concentrated strong acid gives pH < 0, concentrated strong base gives pH > 14

Where to go next: drill numbers in 2.6.11-pH-calculationsfor-strong-acids-bases and 2.6.12-pH-calculations-for-weak-acids-bases; see the same log-trick applied to precipitation in 3.4.5-Solubility-product-constant. Foundational recap: 2.6.1-Arrhenius-acids-and-bases.