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Ask for the pH of 1.0 × 10⁻⁸ M hydrochloric acid and many students write 8 at once. That answer says that adding acid to water made it basic, which cannot be right. The mistake is ignoring the H⁺ that water itself provides. This post explains why that matters at very low concentrations, gives one exact formula that always works, shows exactly when you may drop water's contribution, and works four cases with numbers checked to two decimal places.
Where is this in NCERT and the syllabus?
The current NCERT Class 11 Chemistry Part I chapter Equilibrium (2026-27 reprint) defines the ionic product of water, Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K (Section 6.11.1), and notes that Kw depends on temperature. Its Problem 6.17 asks for the pH of 1.0 × 10⁻⁸ M HCl and solves x² + 10⁻⁸x − 10⁻¹⁴ = 0 to get pH 6.98. The JEE (Main) 2026 syllabus lists "ionization of water, pH scale" under ionic equilibrium, and the NEET (UG) 2026 syllabus has the same line. Check your exam year's syllabus.
For weak acids, buffers and salt hydrolysis, see our guide to pH, buffers and ionic equilibrium. This post stays with the very dilute strong acid and turns it into a method you can reuse.
Why does water's H⁺ matter here?
Pure water at 298 K has [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ M. In 0.01 M HCl the acid supplies 10⁻² M H⁺, a hundred thousand times more than water, so water's share is lost in the rounding. In 10⁻⁸ M HCl the acid supplies only a tenth of what pure water does. Water's contribution is now the larger part, and leaving it out is not an approximation; it is wrong.
The added acid does push water's own ionisation back, by Le Chatelier's principle, so water supplies a little less than 10⁻⁷ M. That is why you need an equation rather than simply adding 10⁻⁸ and 10⁻⁷.
What is the exact method?
Let the acid concentration be c and let water supply x mol/L of H⁺ and the same amount of OH⁻.
[H⁺] = c + x and [OH⁻] = x.
Kw = (c + x)x, so x² + cx − Kw = 0.
Solve the quadratic and take the positive root.
Writing h = c + x for the total [H⁺] gives one formula you can use for any strong acid concentration:
h = c/2 + √(c²/4 + Kw)
When c is large, this reduces to h ≈ c. When c is tiny, it reduces to h ≈ √Kw = 10⁻⁷ M, pure water. It never gives a pH above 7 for an acid.
Worked cases
All at 298 K with Kw = 1.0 × 10⁻¹⁴.
Case 1: 1.0 × 10⁻⁸ M HCl. h = 0.5 × 10⁻⁸ + √(0.25 × 10⁻¹⁶ + 10⁻¹⁴) = 1.05 × 10⁻⁷ M. pH = 6.98, matching NCERT. Water supplies x = 9.5 × 10⁻⁸ M, more than nine times the acid's share.
Case 2: 1.0 × 10⁻⁷ M HCl. The shortcut gives pH 7.00, a neutral acid. The formula gives h = 0.5 × 10⁻⁷ + √(0.25 × 10⁻¹⁴ + 10⁻¹⁴) = 1.62 × 10⁻⁷ M, so pH = 6.79.
Case 3: 1.0 × 10⁻⁶ M HCl. h = 1.0099 × 10⁻⁶ M, so pH = 5.996, which rounds to 6.00. Here the shortcut is fine: water adds about 1% to [H⁺].
Case 4: 1.0 × 10⁻⁸ M NaOH. The same algebra applies to OH⁻: [OH⁻] = 1.05 × 10⁻⁷ M, pOH = 6.98 and pH = 7.02. Slightly basic, as it should be.
A common exam version: an HCl solution of pH 5 is diluted 1000 times. The new concentration is 10⁻⁸ M, so the pH is 6.98, not 8.
When can you ignore water?
From the cases above, a working rule at 298 K:
c of 10⁻⁶ M or more: use pH = −log c. The error is under 0.01 pH unit.
c between about 10⁻⁶ and 10⁻⁸ M: use the quadratic.
c far below 10⁻⁸ M: the pH approaches 7.00 and the acid makes almost no difference.
Which mistakes cost marks?
Writing pH = 8 for 10⁻⁸ M acid. Always run the sense check.
Adding 10⁻⁷ and 10⁻⁸ directly. That gives 1.1 × 10⁻⁷ M and pH 6.96. Close, but the method is wrong, because added acid suppresses water's ionisation.
Taking the negative root. A concentration cannot be negative.
Using Kw = 10⁻¹⁴ at other temperatures. NCERT notes Kw changes with temperature; use the value a question gives.
Applying this to weak acids without thought. A very dilute weak acid needs both its Ka and Kw. Start with the strong-acid case before trying that.
If you lose track of which approximation you made, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can see where a step assumes water is negligible.
A practice routine
Find the pH of 5.0 × 10⁻⁹ M HCl. Check: about 6.99.
Find the pH of 2.0 × 10⁻⁷ M NaOH. Check: [OH⁻] = 2.41 × 10⁻⁷ M and pH ≈ 7.38.
Derive h = c/2 + √(c²/4 + Kw) from Kw = (c + x)x without notes.
Mix these with three ordinary strong-acid questions and decide, before calculating, whether water matters in each.
Tonight, redo case 2 from a blank page and confirm that 10⁻⁷ M HCl has pH 6.79.
