pH of Very Dilute Acids: Why 10⁻⁸ M HCl Is Not pH 8

By Aniket Bhardwaj · 13 October 2026 · Updated 13 October 2026 · Class 11 Chemistry

Ask a Class 11 student for the pH of 10⁻⁸ M HCl and most will answer 8. That answer is wrong, and it is wrong in a way that board examiners like to test. A strong acid can never give a solution with pH above 7 at 25 °C. The mistake comes from forgetting that water itself supplies H⁺ ions. This page shows you when that matters and how to get the correct answer.

A comparison of five dilute HCl concentrations shows shortcut pH values rising from 6.00 to 8.00 while correct pH values stay below 7, from 5.996 to 6.98. — Water keeps very dilute HCl acidic: its pH approaches 7 from below.

Why the simple formula fails

For a normal acid solution, pH = −log[H⁺], and [H⁺] is just the acid concentration. This works because the H⁺ from water (about 10⁻⁷ M in pure water at 25 °C) is tiny next to the acid. When the acid concentration falls to about 10⁻⁶ M or lower, the acid and the water give H⁺ in comparable amounts, and you must count both.

K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ (at 25 °C)
Charge balance for a strong monoprotic acid of concentration C: [H⁺] = C + [OH⁻]
Substitute [OH⁻] = K_w/[H⁺] and rearrange:
[H⁺]² − C[H⁺] − K_w = 0
[H⁺] = ( C + √(C² + 4K_w) ) / 2

What each symbol means

C is the molar concentration of the strong acid you added (mol/L). A strong monoprotic acid such as HCl or HNO₃ is taken as fully dissociated, so it gives C mol/L of H⁺.
[H⁺] is the total hydrogen-ion concentration, from the acid and from water together.
[OH⁻] is the hydroxide concentration. In an acid solution every OH⁻ comes from water, so it also tells you how much H⁺ water supplied.
K_w is the ionic product of water. It is 1.0 × 10⁻¹⁴ at 25 °C and changes with temperature. All the numbers below assume 25 °C.

Worked examples

Example 1: pH of 1.0 × 10⁻⁸ M HCl.
Step 1: C = 1.0 × 10⁻⁸ and K_w = 1.0 × 10⁻¹⁴.
Step 2: C² = 1.0 × 10⁻¹⁶. 4K_w = 4.0 × 10⁻¹⁴. The sum is 4.001 × 10⁻¹⁴, and its square root is 2.0025 × 10⁻⁷.
Step 3: [H⁺] = (1.0 × 10⁻⁸ + 2.0025 × 10⁻⁷) / 2 = 2.1025 × 10⁻⁷ / 2 = 1.0513 × 10⁻⁷ M.
Step 4: pH = −log(1.0513 × 10⁻⁷) = 6.978.
Check: [OH⁻] = 10⁻¹⁴ / 1.0513 × 10⁻⁷ = 9.51 × 10⁻⁸. Then C + [OH⁻] = 0.1 × 10⁻⁷ + 0.951 × 10⁻⁷ = 1.051 × 10⁻⁷ ✓.
Answer: pH ≈ 6.98, slightly acidic. Almost all the H⁺ (about 90%) comes from water, not from the HCl.

Example 2: pH of 1.0 × 10⁻⁷ M HCl.
Step 1: C² = 1.0 × 10⁻¹⁴, 4K_w = 4.0 × 10⁻¹⁴, sum = 5.0 × 10⁻¹⁴, square root = 2.236 × 10⁻⁷.
Step 2: [H⁺] = (1.0 × 10⁻⁷ + 2.236 × 10⁻⁷) / 2 = 1.618 × 10⁻⁷ M.
Step 3: pH = −log(1.618 × 10⁻⁷) = 6.79.
Answer: pH ≈ 6.79, not 7. The common wrong answer "pH = 7" would make an acid neutral.

Example 3: when the shortcut is good enough: 1.0 × 10⁻⁶ M HCl.
Step 1: C² = 1.0 × 10⁻¹², 4K_w = 4.0 × 10⁻¹⁴, sum = 1.04 × 10⁻¹², square root = 1.0198 × 10⁻⁶.
Step 2: [H⁺] = (1.0 × 10⁻⁶ + 1.0198 × 10⁻⁶) / 2 = 1.0099 × 10⁻⁶ M.
Step 3: pH = 5.9957.
Answer: pH = 5.996, against 6.000 from the shortcut. The difference is only 0.004, so for 10⁻⁶ M and stronger acids you can use pH = −log C.

Example 4: a very dilute base, 1.0 × 10⁻⁷ M NaOH.
Step 1: The same algebra works with the roles swapped: [OH⁻] = ( C + √(C² + 4K_w) ) / 2 = 1.618 × 10⁻⁷ M.
Step 2: pOH = 6.79.
Step 3: pH = 14 − 6.79 = 7.21.
Answer: pH ≈ 7.21, basic but only slightly. The shortcut (pOH = 7, pH = 7) would call it neutral.

Quick comparison table

HCl concentration (M)	Shortcut pH = −log C	Correct pH (with water)
1.0 × 10⁻⁶	6.00	5.996
1.0 × 10⁻⁷	7.00	6.79
5.0 × 10⁻⁸	7.30	6.89
2.0 × 10⁻⁸	7.70	6.96
1.0 × 10⁻⁸	8.00	6.98

Notice the trend. As the acid gets more and more dilute, the correct pH does not climb past 7. It creeps towards 7 from below, because the solution looks more and more like pure water.

Reporting pH 8 for 10⁻⁸ M HCl. An acid cannot make a solution basic. If your answer is above 7 for an acid, you have made an error.
Adding the two concentrations blindly. The simple sum C + 10⁻⁷ overcounts water's contribution, because adding acid pushes the water equilibrium back. In Example 1, naive addition gives 1.1 × 10⁻⁷ (pH 6.96), while the correct value is 1.05 × 10⁻⁷ (pH 6.98). Use the quadratic.
Using the quadratic for strong solutions. For 0.1 M HCl the water term is negligible. Using pH = −log C is correct and faster.
Forgetting the temperature. K_w = 10⁻¹⁴ is a 25 °C value. If a question gives another K_w, use that value in the quadratic.
Applying this to weak acids. A weak acid needs its K_a. The method above is only for strong acids.

When does it matter in exams?

Situation	What to do
Strong acid, C ≥ 10⁻⁶ M	Use pH = −log C
Strong acid, C near or below 10⁻⁷ M	Use [H⁺] = ( C + √(C² + 4K_w) ) / 2
Question says "pH of 10⁻⁸ M HCl"	Expect a value just under 7, and say that water's H⁺ must be counted
Dilute strong base	Same quadratic for [OH⁻], then pH = 14 − pOH

Different boards and entrance exams treat this topic with different depth. Check your own syllabus before spending a lot of time on it.

Work the steps by hand first. Then use the calculator to check the pH of ordinary-strength solutions.

Open the pH / pOH Calculator →