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pH Calculator

Calculate pH, pOH, [H⁺] and [OH⁻] from any one of them, or from a strong acid, strong base, weak acid (Ka) or weak base (Kb) concentration — solved exactly, with steps. Free for AP Chemistry and A-level.

pH (acidic)

2.88
pOH 11.12[H+]=1.33×10−3[\mathrm{H^+}] = 1.33\times10^{-3}[OH−]=7.50×10−12[\mathrm{OH^-}] = 7.50\times10^{-12}
0 · acidic7 · neutralbasic · 14
  1. 1

    Equilibrium: K_a = x² / (C − x)

    Solved exactly with the quadratic formula (no "x is small" approximation).

    x=−Ka+Ka2+4KaC2=−1.80×10−5+(1.80×10−5)2+4(1.80×10−5)(1.00×10−1)2=1.33×10−3 Mx = \frac{-K_a + \sqrt{K_a^2 + 4K_aC}}{2} = \frac{-1.80\times10^{-5} + \sqrt{(1.80\times10^{-5})^2 + 4(1.80\times10^{-5})(1.00\times10^{-1})}}{2} = 1.33\times10^{-3}\ \mathrm{M}
  2. 2

    Percent ionisation

    xC×100=1.33%\frac{x}{C}\times100 = 1.33\%
  3. 3

    pH = −log x

    pH=−log⁡(1.33×10−3)=2.88\text{pH} = -\log(1.33\times10^{-3}) = 2.88
  4. 4

    pOH and [OH⁻] (25 °C: pH + pOH = 14)

    pOH=14−2.88=11.12,[OH−]=10−14[H+]=7.50×10−12 M\text{pOH} = 14 - 2.88 = 11.12,\quad [\mathrm{OH^-}] = \frac{10^{-14}}{[\mathrm{H^+}]} = 7.50\times10^{-12}\ \mathrm{M}

At 25 °C (Kw = 1.0 × 10⁻¹⁴). Polyprotic acids such as H₂SO₄ are treated as fully releasing the number of H⁺ you enter.

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FAQ

How do you calculate pH?

pH = −log[H⁺]. At 25 °C, pH + pOH = 14 and [H⁺][OH⁻] = 1.0 × 10⁻¹⁴.

How is the pH of a weak acid found?

From Ka = x²/(C − x). The calculator solves this quadratic exactly instead of assuming x is small, then pH = −log x.

Why is very dilute HCl not pH 8?

At concentrations below about 10⁻⁶ M, water’s own H⁺ matters. The calculator includes it, so 10⁻⁸ M HCl gives pH ≈ 6.98, not 8.

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