pH, pOH, pKa, pKb scales: Worked examples — Chemistry deep dive

This page is the drill hall for the p-scales topic. The parent note built the definitions; here we throw every possible kind of problem at those definitions until none can surprise you.

Before we start, one promise about notation — every symbol used below was defined in the parent note, but let us re-anchor the three we lean on hardest, so a reader landing here cold is never lost.

The scenario matrix

Every problem this topic can throw at you lives in one of these cells. The examples below are each tagged with the cell they kill.

Cell	Case class	What makes it tricky
A	pH → [H⁺] (forward, non-integer)	inverse log, scientific notation
B	[OH⁻] → pH (cross the pH–pOH bridge)	must go through pOH first
C	Neutral / degenerate: pure water	[H⁺]=[OH⁻], the balance point
D	Limiting: ultra-dilute strong acid	naïve formula gives wrong sign of acidity
E	pKa ↔ pKb conjugate flip	uses the $= 14$ bridge
F	Compare two acids by pKa	ratio = 10^(ΔpKa)
G	pH = pKa buffer midpoint	50/50 point, links to [[2.6.8-Henderson-Hasselbalch-equation
H	Real-world word problem	translate English → concentration
I	Exam twist: mixing / negative pH	pH can go below 0 and above 14

Cell A — pH to [H⁺], non-integer

Worked example Example 1 (Cell A)

Problem: A cola drink measures pH = 2.5. What is $[H^{+}]$ ?

Forecast: Guess first — will the answer be bigger or smaller than $1 0^{- 3}$ M? (Lower pH than 3 means more acid, so bigger concentration.)

Write the definition backwards: $[H^{+}] = 1 0^{- pH}$ . Why this step? The exponential is the exact undo of $- lo g_{10}$ ; if $pH = - lo g_{10} [H^{+}]$ then raising 10 to $- pH$ recovers the original number.
Substitute: $[H^{+}] = 1 0^{- 2.5}$ . Why this step? Just plugging the measured value into the inverse.
Split the exponent: $1 0^{- 2.5} = 1 0^{0.5} \times 1 0^{- 3} = 3.16 \times 1 0^{- 3}$ M. Why this step? Breaking $- 2.5$ into an easy power of ten ( $- 3$ ) times a leftover ( $1 0^{0.5} \approx 3.16$ ) is how you read it without a full calculator.

Verify: Take $- lo g_{10} (3.16 \times 1 0^{- 3})$ . That is $- (lo g 3.16 - 3) = - (0.5 - 3) = 2.5$ . ✓ And $3.16 \times 1 0^{- 3} > 1 0^{- 3}$ , matching our forecast of "more acidic."

Cell B — [OH⁻] to pH, crossing the bridge

Cell C — the neutral, degenerate case

Cell D — the limiting trap: ultra-dilute strong acid

This is the case that catches almost everyone. See strong-acid pH calculations for the full machinery.

Worked example Example 4 (Cell D — limiting behaviour)

Problem: You add HCl until its formal concentration is $1.0 \times 1 0^{- 8}$ M. Naïvely, $pH = - lo g_{10} (1 0^{- 8}) = 8$ . But that says an acid is basic — impossible! Fix it.

Forecast: The true pH must be just below 7 (slightly acidic), never above 7.

Spot the flaw: at $1 0^{- 8}$ M, the acid's $H^{+}$ is smaller than water's own $1 0^{- 7}$ . You cannot ignore water. Why this step? The naïve formula assumes the acid dominates; here it doesn't, so the assumption breaks.
Charge balance: $[H^{+}] = [OH^{-}] + [Cl^{-}]$ , with $[Cl^{-}] = 1 0^{- 8}$ (all HCl dissociated). Why this step? Total positive charge must equal total negative charge — the honest bookkeeping when sources are comparable.
Substitute $[OH^{-}] = K_{w} / [H^{+}]$ : let $h = [H^{+}]$ , then $h = \frac{1 0 ^{- 14}}{h} + 1 0^{- 8}$ . Why this step? $K_{w}$ links the two ions so we get one equation in one unknown.
Rearrange to $h^{2} - 1 0^{- 8} h - 1 0^{- 14} = 0$ and solve the quadratic: $h = \frac{1 0 ^{- 8} + ( 1 0 ^{- 8} ) ^{2} + 4 \cdot 1 0 ^{- 14}}{2} = 1.05 \times 1 0^{- 7} M .$ Why this step? A quadratic is the tool because two additive sources of $H^{+}$ multiply into a squared term; only the positive root is physical (concentration can't be negative).
$pH = - lo g_{10} (1.05 \times 1 0^{- 7}) = 6.98$ .

Verify: $6.98 < 7$ ✓ — genuinely acidic, forecast confirmed. Contrast the naïve "8." Check $K_{w}$ : $[OH^{-}] = 1 0^{- 14} /1.05 \times 1 0^{- 7} = 9.5 \times 1 0^{- 8}$ ; then $h - [OH^{-}] = 1.05 \times 1 0^{- 7} - 9.5 \times 1 0^{- 8} = 1.0 \times 1 0^{- 8} = [Cl^{-}]$ ✓.

Cell E — conjugate flip pKa ↔ pKb

Cell F — comparing two acids by pKa

Cell G — the buffer midpoint pH = pKa (with figure)

Worked example Example 7 (Cell G — geometric)

Problem: A buffer is made from an acid with $pK_{a} = 4.76$ so that $[A^{-}] = [HA]$ . What is the pH? Show why geometrically.

Forecast: Equal amounts of acid and conjugate base — the "balanced seesaw." pH should land exactly on pKa.

Start from Henderson–Hasselbalch: $pH = pK_{a} + lo g_{10} \frac{[ A ^{-} ]}{[ HA ]}$ . Why this step? This equation is the pKa/pH relationship rearranged for buffers (buffers).
Set the ratio to 1: $lo g_{10} (1) = 0$ . Why this step? When the two are equal, their ratio is exactly 1, and $lo g_{10} 1 = 0$ — the log of "no imbalance."
Therefore $pH = pK_{a} + 0 = 4.76$ .

Look at the figure: the curve is the fraction of acid dissociated as pH rises. At the flat, steepest-resistance crossover, the two coloured areas (HA and A⁻) are equal, and that point sits directly above $pH = pK_{a}$ .

Verify: $lo g_{10} (1) = 0$ so $pH = pK_{a}$ exactly ✓. This is the maximum-buffering midpoint — 50/50 as the parent note proved.

Cell H — real-world word problem

Cell I — exam twist: negative pH and beyond the 0–14 range

Worked example Example 9 (Cell I — degenerate/limiting extreme)

Problem: Concentrated battery acid is effectively $[H^{+}] = 5.0$ M (strong acid, fully dissociated). Find pH. Then a lab base has $[OH^{-}] = 3.0$ M — find its pH.

Forecast: These concentrations are above 1 M, so the logs flip sign — pH will fall below 0 for the acid and rise above 14 for the base. The "0–14 scale" is only a convenience, not a law.

Acid part:

$pH = - lo g_{10} (5.0) = - 0.70$ . Why this step? $lo g_{10}$ of a number bigger than 1 is positive, so its negative is below zero — perfectly valid.

Base part: 2. $pOH = - lo g_{10} (3.0) = - 0.48$ . Why this step? Same reasoning applied to hydroxide. 3. $pH = 14 - (- 0.48) = 14.48$ . Why this step? The pH–pOH bridge still holds; it just pushes us past 14.

Verify: $[H^{+}] = 1 0^{- (- 0.70)} = 1 0^{0.70} = 5.0$ M ✓. For the base, $1 0^{- (- 0.48)} = 3.0$ M of $OH^{-}$ ✓. Negative and >14 pH values are real and expected for concentrated solutions.

Wrap-up recall

Recall Which cell is each phrase pointing at?

"Ultra-dilute HCl needs a quadratic" ::: Cell D — water's own ions matter. "pH = pKa" ::: Cell G — the 50/50 buffer midpoint. "ΔpKa of 2 means 100× stronger" ::: Cell F — ratio is $1 0^{Δ pK_{a}}$ . "pH can be −0.7" ::: Cell I — concentrations above 1 M break the 0–14 comfort zone. "Given [OH⁻], go through pOH then subtract from 14" ::: Cell B — cross the bridge.