How to Calculate pH From Concentration

pH = −log [H+]

[H+] is in mol/L. The logarithm is base 10. A tenfold drop in [H+] raises pH by 1.

The pH calculator takes a concentration you provide and computes that −log. No account. If you type the formal concentration of acetic acid, you will get the strong-acid number, which is the wrong pH. The tool does not know Ka.

Strong acids: [H+] ≈ the molarity

HCl, HBr, HI, HNO3, and the first proton of H2SO4 are the usual “strong” list. In water they are treated as fully dissociated.

For 0.025 M HCl:

[H+] = 0.025 mol/L
pH = −log(0.025) = 1.60

Check on a calculator: log(0.025) = log(2.5 × 10⁻²) = log 2.5 + log 10⁻² ≈ 0.40 − 2 = −1.60, then the minus sign makes +1.60.

0.025 M is 2.5 × 10⁻² M, not 0.25 M and not 25 mM mis-entered as 25 M. 25 mM is 0.025 M. If you enter 25 and a unit of M, pH goes negative and you should stop.

H2SO4 is diprotic and strong on the first proton; the second is weaker. Many first-year problems still say “0.025 M H2SO4, assume both protons” and want [H+] = 0.050 M, pH = 1.30. That assumption is already an approximation. Read the problem.

Strong bases, pOH, and 14 at 25 °C

NaOH and KOH: [OH−] ≈ formal molarity.

0.025 M NaOH:
pOH = −log(0.025) = 1.60
pH = 14.00 − 1.60 = 12.40 (25 °C)

Kw = [H+][OH−] = 1.0 × 10⁻¹⁴ at 25 °C, so pH + pOH = 14.00. At 0 °C or 50 °C, Kw is different; using 14 is then a 25 °C habit, not a law of nature.

If you report pH = 1.60 for NaOH, you computed pOH and labeled it pH. That is a full-credit-to-zero error on some rubrics.

Weak acids: this formula is not enough

Acetic acid, HF, formic acid, ammonium ion - [H+] is not equal to C.

The usual first approximation for a weak acid HA is

[H+] ≈ √(Ka C) when C / Ka is large enough (often C/Ka > 100 as a rule of thumb)

For C = 0.025 M, Ka = 1.8 × 10⁻⁵:

[H+] ≈ √(4.5 × 10⁻⁷) ≈ 6.7 × 10⁻⁴ M
pH ≈ 3.17

Compare to 1.60 if you had used −log(0.025). Almost two pH units is not “close enough.”

If C/Ka is not large, solve x² / (C − x) = Ka. Buffers are a different equation (Henderson–Hasselbalch) and a different lab.

The pH calculator will not run Ka for you. Do not paste 0.025 M acetic acid into it and trust the output.

Very dilute strong acids and water

Pure water at 25 °C has [H+] = 1.0 × 10⁻⁷ M, pH 7.00.

A “10⁻⁸ M HCl” problem is famous: −log(10⁻⁸) = 8, which would be basic. That is wrong. The acid adds 1.0 × 10⁻⁸ M H+, but water still contributes. The charge-balance / Kw solution gives pH slightly below 7 (about 6.98), not 8.

Practical cutoff: if the strong-acid concentration is within a couple of orders of 10⁻⁷ M, include water. For 0.025 M HCl, water’s 10⁻⁷ is a 0.0004% correction. Ignore it.

Concentration units before the log

pH = −log[H+] needs [H+] in mol/L.

• 25 mM = 0.025 M
• 250 μM = 2.50 × 10⁻⁴ M → pH 3.60 if that is [H+]
• 10% w/w HCl is not 10 M and is not pH −1 from −log(10)

Convert with the concentration calculator or molarity first if you were given grams or percent. Diluting a stock changes C; use C1V1 = C2V2 before you take the log.

pH is dimensionless (it is a log of a concentration relative to 1 M in the intro definition). Do not write “pH = 1.60 M.”

What the local tool will and will not do

Enter [H+] in the pH calculator when you already have that number - typically a strong acid, or [H+] from a completed equilibrium. For a weak acid, finish the Ka algebra on paper, then take −log of the x you found.

If the next step is making 0.025 M from 1.0 M stock, that is dilution, not a new pH formula. After the new C exists, then pH.