pH Calculation Techniques
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Direct answer
A pH numerical is never open-ended: it is one of six templates. Strong acids and bases take total [H+] directly (pH = −log C); weak electrolytes use [H+] = √(Ka C), giving pH = 1/2(pKa − log C); buffers take Henderson-Hasselbalch pH = pKa + log([salt]/[acid]); mixtures are resolved by moles before logs; salts are placed by hydrolysis type; dilution shifts pH by the log of the factor — 1 unit per tenfold dilution for a strong acid, only 0.5 for a weak one. At 298 K, pKw = 14 anchors every conversion between pH and pOH.
What you must remember
- Strong systems: pH = −log C works down to about 10^-6 M; below that, water's 10^-7 M contribution matters — the famous 10^-8 M HCl answer is pH ≈ 6.98, never 8.
- Weak acid: Ka = Cα^2 with α small gives [H+] = √(Ka C) and pH = 1/2(pKa − log C); weak base pOH = 1/2(pKb − log C). The approximation is safe when C/Ka exceeds roughly 100.
- Buffer: pH = pKa + log([salt]/[acid]), reliable within pKa ± 1; buffer capacity is maximal when salt equals acid.
- Dilution arithmetic: tenfold dilution raises a strong acid's pH by 1 but a weak acid's by only 0.5, because log C drops by 1 and the half-factor halves it.
- Mixtures: strong plus strong means total moles of H+ over total volume; a strong acid plus a weak acid is governed by the strong one, since the weak dissociation is further suppressed.
- Salt hydrolysis at 298 K: SA/SB neutral; WA/SB (sodium acetate) pH = 7 + 1/2(pKa + log C); SA/WB (ammonium chloride) pH = 7 − 1/2(pKb + log C); WA/WB salt pH = 7 + 1/2(pKa − pKb), independent of concentration.
- Polyprotic acids: the first Ka dominates; [H+] ≈ √(Ka1 C) for carbonic or sulphurous acid systems at moderate dilution.
- Temperature: Kw rises with temperature, so pure water's pH falls below 7 on heating while the solution stays exactly neutral.
A pH session from weak acid to buffer
Work 0.1 M acetic acid, pKa 4.74: pH = 1/2(4.74 − log 0.1) = 1/2(4.74 + 1) = 2.87. Dilute a hundredfold: pH = 1/2(4.74 + 3) = 3.87 — exactly one unit up, while a strong acid would have gained two. Now build a buffer: 0.15 M sodium acetate plus 0.05 M acetic acid gives pH = 4.74 + log 3 = 5.22. Add one millimole of HCl per litre: it converts acetate to acid, the ratio moves to 0.149/0.051, and pH barely flickers to 5.20 — the entire point of buffer arithmetic. Finish with ammonium chloride: pKb of ammonia is 4.74 and C = 0.1 M, so pH = 7 − 1/2(4.74 + (−1)) = 5.13, mildly acidic from the conjugate acid's hydrolysis. Four template calls, four answers, and nothing memorised beyond the templates themselves.
Where marks are lost
The recurring JEE Main errors are mechanical. Mixing two acid solutions without halving concentrations for equal volumes turns pH 2.5 into 2.2. Averaging pH values instead of adding [H+] concentrations before the logarithm. Using molarities where the Henderson equation actually wants the mole ratio (volumes cancel, so moles work directly). JEE Advanced adds layered cases: a weak acid partly neutralised by strong base leaves a buffer whose composition is fixed by reaction stoichiometry, or a WA/WB salt whose pH ignores concentration entirely. The assertion-reason classic remains the dilute strong acid: "the pH of 10^-8 M HCl is 8" is false because water autoionisation supplies nearly a hundred times more H+ than the acid, and the acid merely shifts [H+] from 10^-7 to about 1.05 × 10^-7 M.
Frequently asked questions
What is the pH of 10^-8 M HCl?
About 6.98 — water's autoionisation supplies nearly all the hydrogen ion, so the acid slightly raises [H+] above 10^-7 M and the solution stays acidic.
How does the pH of a weak acid change on hundredfold dilution?
It rises by exactly one unit, 0.5 per tenfold step, from pH = 1/2(pKa − log C), until the approximation itself fails near millimolar concentrations.
When does the Henderson-Hasselbalch equation fail?
Outside the pKa ± 1 window, at extreme dilution, or when the acid dissociates appreciably — the equation assumes both species remain essentially undissociated.
What determines the pH of an ammonium chloride solution?
Hydrolysis of the ammonium ion as a conjugate acid: pH = 7 − 1/2(pKb + log C) at 298 K, so 0.1 M NH4Cl sits near 5.1.
Which acid fixes the pH of a strong-plus-weak acid mixture?
The strong acid's hydrogen ions suppress the weak acid's already small dissociation, so the pH is essentially that of the strong acid alone.