PH and ionic product

Drop a strip of litmus into lemon juice and it turns red; drop it into soap solution and it turns blue. Behind that simple colour change sits a precise number line — the pH scale — built entirely on a fact that sounds strange at first: even pure water breaks apart, ever so slightly, into H+H^+ and OH−OH^- ions. This self-ionisation of water fixes a constant, the ionic product of water KwK_w, that silently governs every aqueous acid, base and salt solution you will ever calculate a pH for. This article takes that one constant and builds outward: how KwK_w gives the neutral point of pH=7\text{pH} = 7, how pH+pOH=14\text{pH} + \text{pOH} = 14 falls straight out of it, how pKa\text{p}K_a and pKb\text{p}K_b extend the same logic to acid and base strength, and how you actually compute pH for a strong acid versus a weak acid — two genuinely different calculations that examiners love to mix up.

The core insight

Water's self-ionisation fixes Kw=[H+][OH−]=1.0×10−14K_w = [H^+][OH^-] = 1.0\times10^{-14} at 298 K, and taking −log⁡10-\log_{10} of it gives the master relation pH+pOH=14\text{pH} + \text{pOH} = 14 that underlies every pH calculation.

The core idea

Water is not perfectly inert — a tiny fraction of molecules ionise, or more precisely transfer a proton between two water molecules:

2H2O⇌H3O++OH−2H_2O \rightleftharpoons H_3O^+ + OH^-

This is usually written in the shorthand H2O⇌H++OH−H_2O \rightleftharpoons H^+ + OH^-. Since the extent of ionisation is so small that the concentration of un-ionised water barely changes, the equilibrium constant for this process is folded into a single number, the ionic product of water:

Kw=[H+][OH−]K_w = [H^+][OH^-]

where: [H+][H^+] and [OH−][OH^-] are the molar concentrations of hydrogen and hydroxide ions in mol/L, and KwK_w has units mol2L−2\text{mol}^2\text{L}^{-2}. At 298 K, Kw=1.0×10−14K_w = 1.0\times10^{-14}, and because pure water produces H+H^+ and OH−OH^- in a strict 1:1 ratio, [H+]=[OH−]=1.0×10−7 M[H^+] = [OH^-] = 1.0\times10^{-7}\ M — the neutral condition. KwK_w is not a universal constant like π\pi; it changes with temperature because self-ionisation is an endothermic process, so heating water increases KwK_w and shifts the neutral point away from 10−7 M10^{-7}\ M. Nearly all ISC-level problems fix the temperature at 298 K, so this article does too unless stated otherwise.

The pH scale compresses this tiny, exponentially varying concentration into a small, manageable number:

pH=−log⁡10[H+]\text{pH} = -\log_{10}[H^+]

and, mirroring it, pOH=−log⁡10[OH−]\text{pOH} = -\log_{10}[OH^-]. A neutral solution sits at pH=7\text{pH} = 7; an acidic solution has [H+]>10−7 M[H^+] > 10^{-7}\ M so pH<7\text{pH} < 7; a basic solution has [H+]<10−7 M[H^+] < 10^{-7}\ M so pH>7\text{pH} > 7. The same logarithmic idea extends to the ionisation constants of weak acids and bases: pKa=−log⁡10Ka\text{p}K_a = -\log_{10}K_a and pKb=−log⁡10Kb\text{p}K_b = -\log_{10}K_b. A smaller pKa\text{p}K_a means a larger KaK_a, i.e. a stronger acid — the scale runs opposite to intuition at first glance, which is exactly why it trips students up. For any conjugate acid–base pair (an acid and the base left behind when it donates a proton), the two constants are linked through water's own equilibrium:

Ka×Kb=Kw  ⟹  pKa+pKb=pKw=14 (at 298 K)K_a \times K_b = K_w \quad\implies\quad \text{p}K_a + \text{p}K_b = \text{p}K_w = 14 \ \text{(at 298 K)}

where: KaK_a is the ionisation constant of the acid, KbK_b that of its conjugate base. A strong conjugate acid (large KaK_a) automatically has a very weak conjugate base (tiny KbK_b), and vice versa — the two cannot both be strong. Two distinct calculation routes follow from all this. For a strong acid or base, ionisation is essentially complete, so [H+][H^+] (or [OH−][OH^-]) is read directly off the stated concentration. For a weak acid or base, only a small fraction ionises, so [H+][H^+] must first be worked out from KaK_a using an ICE table (Ostwald's dilution law), and only then converted to pH. Mixing up these two routes is the single most common calculation error at this level. Acid–base indicators are themselves weak acids or bases whose ionised and un-ionised forms are different colours; each indicator changes colour over a pH range centred near its own pKa\text{p}K_a (or pKIn\text{p}K_{In}), which is why choosing the right indicator for a titration means matching that range to the sharp pH jump near the equivalence point.

Figure
Number line from pH 0 to pH 14 showing acidic basic and neutral regions with the hydrogen ion concentration marked at each end
The pH scale at 298 K acidic — pH less than 7 neutral basic — pH greater than 7 pH 0 [H+] = 1 M pH 7 [H+] = 10 to the minus 7 M pH 14 [H+] = 10 to the minus 14 M Each unit of pH is a tenfold change in hydrogen ion concentration
Derivation AFrom the ionic product of water to pH plus pOH equals 14

The goal is to show, using nothing more than the definition of a logarithm, that the two everyday numbers pH and pOH of any aqueous solution at 298 K are not independent quantities but are locked together by pH+pOH=14\text{pH} + \text{pOH} = 14, and that this single relation is the reason a rise in [H+][H^+] always forces a matching fall in [OH−][OH^-].

Step 1. Start from the defining equilibrium expression for the ionic product of water, Kw=[H+][OH−]K_w = [H^+][OH^-], and insert its known value at 298 K: [H+][OH−]=1.0×10−14[H^+][OH^-] = 1.0\times10^{-14}.

Step 2. Take −log⁡10-\log_{10} of both sides. The left side splits using the logarithm rule for a product, log⁡10(ab)=log⁡10a+log⁡10b\log_{10}(ab) = \log_{10}a + \log_{10}b, so −log⁡10[H+]−log⁡10[OH−]=−log⁡10(1.0×10−14)-\log_{10}[H^+] - \log_{10}[OH^-] = -\log_{10}(1.0\times10^{-14}).

Step 3. Recognise each term by its definition: −log⁡10[H+]-\log_{10}[H^+] is exactly pH\text{pH}, and −log⁡10[OH−]-\log_{10}[OH^-] is exactly pOH\text{pOH}. The right-hand side evaluates cleanly because 1.0×10−141.0\times10^{-14} is a power of ten: −log⁡10(10−14)=14-\log_{10}(10^{-14}) = 14.

Step 4. Combine the pieces to read off the result: pH+pOH=14\text{pH} + \text{pOH} = 14 at 298 K. Because this sum is fixed, pH and pOH move in lockstep — if a solution's [H+][H^+] rises (pH falls), [OH−][OH^-] must fall by exactly enough to keep [H+][OH−]=Kw[H^+][OH^-] = K_w, so pOH rises by the same amount pH fell. Knowing either one, and the temperature, always hands you the other.

Worked example

pH of a strong acid, then check with KwK_w. Calculate the pH and pOH of a 0.010 M0.010\ M solution of HCl\text{HCl} at 298 K, and verify the answer against KwK_w.

Step 1. HCl\text{HCl} is a strong acid and ionises essentially completely: HCl→H++Cl−\text{HCl} \rightarrow H^+ + Cl^-. Since each formula unit releases one H+H^+, the concentration is read directly off the given value: [H+]=0.010 M[H^+] = 0.010\ M.

Step 2. Apply the pH definition: pH=−log⁡10[H+]=−log⁡10(1.0×10−2)=2.0\text{pH} = -\log_{10}[H^+] = -\log_{10}(1.0\times10^{-2}) = 2.0.

Step 3. Find pOH using the relation just derived: pOH=14−pH=14−2.0=12.0\text{pOH} = 14 - \text{pH} = 14 - 2.0 = 12.0, so [OH−]=10−12 M[OH^-] = 10^{-12}\ M — a vanishingly small but strictly non-zero hydroxide concentration, present in every acidic solution.

Step 4. Check for consistency: [H+][OH−]=(10−2)(10−12)=10−14=Kw[H^+][OH^-] = (10^{-2})(10^{-12}) = 10^{-14} = K_w, confirming the calculation is internally consistent.

Answer: pH = 2.0, pOH = 12.0.

Competitive stretch

For a weak acid the same target concentration gives a very different, higher pH because ionisation is only partial. Taking 0.050 M0.050\ M acetic acid with Ka=1.8×10−5K_a = 1.8\times10^{-5}: set x=[H+]x = [H^+] at equilibrium and use the small-xx approximation Ka≈x2/cK_a \approx x^2/c, so x2=(1.8×10−5)(0.050)=9.0×10−7x^2 = (1.8\times10^{-5})(0.050) = 9.0\times10^{-7}, giving x≈9.49×10−4 Mx \approx 9.49\times10^{-4}\ M (the approximation is valid since x/c≈1.9%x/c \approx 1.9\%, well under the usual 5% cutoff). Then pH=−log⁡10(9.49×10−4)≈3.02\text{pH} = -\log_{10}(9.49\times10^{-4}) \approx 3.02 — noticeably higher (less acidic) than the strong-acid pH of 2.0 at a similar order of concentration, a direct numerical illustration of why "weak" acids give weaker [H+][H^+].

Common traps

  • Treating KwK_w as a fixed universal constant. Kw=1.0×10−14K_w = 1.0\times10^{-14} and neutral pH=7\text{pH} = 7 hold specifically at 298 K; KwK_w rises with temperature since self-ionisation is endothermic, though ISC problems generally assume 298 K unless stated otherwise.
  • Using the strong-acid shortcut on a weak acid. A weak acid does not ionise completely, so [H+]≠c[H^+] \ne c — it must be computed from KaK_a via an ICE table or Ostwald's dilution law first.
  • Dropping the negative sign or mishandling powers of ten. pH=−log⁡10[H+]\text{pH} = -\log_{10}[H^+]; forgetting the minus sign, or misreading 10−n10^{-n} inside the log, is the most common arithmetic slip.
  • Forgetting pKa\text{p}K_a runs opposite to acid strength. A smaller pKa\text{p}K_a means a larger KaK_a and a stronger acid — students often assume "bigger number, stronger acid," which is backwards.

Ask the tutor: Want a few more pH problems mixing strong and weak acids and bases, with your steps checked as you go?

Why does pure water have a pH of 7 and not 0?

A pH of 0 would mean [H+]=1 M[H^+] = 1\ M, an extremely concentrated acid. Pure water is not acidic at all — it only self-ionises to a tiny extent, giving [H+]=[OH−]=1.0×10−7 M[H^+] = [OH^-] = 1.0\times10^{-7}\ M at 298 K. Since [H+][H^+] equals [OH−][OH^-], water is exactly balanced between acidic and basic, which is what "neutral" means, and −log⁡10(10−7)=7-\log_{10}(10^{-7}) = 7 lands neutral squarely in the middle of the 0–14 scale.

If Kw changes with temperature, does neutral pH always stay 7?

No. pH=7\text{pH} = 7 is neutral only at 298 K, where Kw=1.0×10−14K_w = 1.0\times10^{-14} makes [H+]=[OH−]=10−7 M[H^+] = [OH^-] = 10^{-7}\ M. At a higher temperature KwK_w increases, so the neutral concentration of H+H^+ is larger and neutral pH drops below 7 — the solution is still neutral (since [H+]=[OH−][H^+] = [OH^-] still holds), it is just that "neutral" no longer sits at exactly pH 7. ISC questions almost always specify or assume 298 K, so you can use pH 7 as neutral by default.

Why do we need pKa and pKb if Ka and Kb already tell us acid or base strength?

Because KaK_a and KbK_b values are often extremely small numbers written in scientific notation (like 1.8×10−51.8\times10^{-5}), which are awkward to compare or write repeatedly. Taking −log⁡10-\log_{10} turns them into ordinary small numbers, just like pH does for [H+][H^+]. The tradeoff is that the scale flips direction: since a larger KaK_a becomes a smaller pKa\text{p}K_a after the negative log, stronger acids get smaller pKa\text{p}K_a values, not larger ones — worth double-checking every time you compare two acids by their pKa\text{p}K_a.

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