pH calculator for acids, bases and buffers

Calculate pH, pOH, [H⁺] and [OH⁻] from a concentration, a strong acid or base, a weak acid's Ka or pKa, or a Henderson–Hasselbalch buffer.

更新于 已验证的示例:8

= 0.000018
试一试
pH
pH: 2.88
小数位数:2;取最近值,等距时取偶数末位
pOH
11.12
[H⁺]
0.00133267mol/L
[OH⁻]
7.50373 × 10⁻¹²mol/L
Solution is
Acidic
pH with the √(Ka·C) shortcut
2.872
Shortcut error in [H⁺]
0.67%
Fraction ionised
1.33%
Uses Kw = 1.0 × 10⁻¹⁴ (25 °C) and concentrations in place of activities.

pH 2.88 (acidic): [H⁺] = 0.001333 mol/L and [OH⁻] = 7.504 × 10⁻¹² mol/L; the acid is 1.33% ionised, and the √(Ka·C) shortcut would be off by 0.67%.

pH scale

Strongly acidicAcidicNear neutralBasicStrongly basic03681114neutral, pure waterpH 2.88
计算方法 S
  1. Equilibrium HA ⇌ H⁺ + A⁻

    Ka=x2C−x⇒x2+Kax−KaC=0,Ka=0.000018, C=0.1K_a = \frac{x^2}{C - x} \Rightarrow x^2 + K_a x - K_aC = 0,\quad K_a = 0.000018,\ C = 0.1
  2. Exact root

    x=−Ka+Ka2+4KaC2=0.00133267 mol/Lx = \frac{-K_a + \sqrt{K_a^2 + 4K_aC}}{2} = 0.00133267\ \mathrm{mol/L}

    Water's own ionisation is neglected — fine while [H⁺] ≫ 10⁻⁷ mol/L.

  3. Compare the shortcut

    KaC=0.00134164⇒pH≈2.8724 (0.673% high in [H+])\sqrt{K_aC} = 0.00134164 \Rightarrow \text{pH} \approx 2.8724\ (0.673\%\ \text{high in }[\mathrm{H^+}])
  4. pOH and [OH⁻] at 25 °C

    pH=2.87528,pOH=14−pH=11.1247,[OH−]=7.50373×10−12 mol/L\text{pH} = 2.87528,\quad \text{pOH} = 14 - \text{pH} = 11.1247,\quad [\mathrm{OH^-}] = 7.50373 \times 10^{-12}\ \mathrm{mol/L}

关于pH calculator for acids, bases and buffers

pH is the negative base-10 logarithm of the hydrogen-ion concentration, pH = −log₁₀[H⁺]. At 25 °C the ion product of water, Kw, is 1.0 × 10⁻¹⁴, so pH + pOH = 14. Strong acids and bases are treated as fully dissociated, with water's own ions included; weak acids are solved exactly from Ka with the quadratic x² + Ka·x − Ka·C = 0; buffers use the Henderson–Hasselbalch equation.

The default, 0.1 mol/L acetic acid with Ka = 1.8 × 10⁻⁵, gives pH 2.88 with 1.33 % of the acid ionised. The √(Ka·C) shortcut taught in class gives pH 2.872, because it overstates [H⁺] by 0.67 %.

Concentrations stand in for activities, which works best for dilute solutions. Results below pH 0 or above 14 are possible for concentrated acids and bases, but there the simple formula is only an estimate.

计算示例

[H⁺] = 0.001 mol/L

Start from
[H⁺] concentration
Concentration
0.001
Concentration unit
mol/L
pH
3.00
pOH
11.00
Solution is
Acidic

核验来源:−log₁₀(10⁻³) = 3; pOH = 14 − 3

pH 7 is neutral

Start from
pH value
pH
7
[H⁺]
1 × 10⁻⁷ mol/L
[OH⁻]
1 × 10⁻⁷ mol/L
Solution is
Neutral

核验来源:10⁻⁷ mol/L at 25 °C (Kw = 10⁻¹⁴)

0.01 M HCl

Start from
Strong acid concentration
Concentration
0.01
Concentration unit
mol/L
H⁺ or OH⁻ released per formula unit
1
pH
2.00

核验来源:Python 3.8 decimal: [H⁺] = (C + √(C² + 4Kw))/2 = 0.010000000001 ⇒ pH 2.0000

Very dilute HCl (1e-8 M) is still acidic

Start from
Strong acid concentration
Concentration
1e-8
Concentration unit
mol/L
H⁺ or OH⁻ released per formula unit
1
pH
6.98
[H⁺]
1.05125 × 10⁻⁷ mol/L

核验来源:Python 3.8 decimal: [H⁺] = (10⁻⁸ + √(10⁻¹⁶ + 4×10⁻¹⁴))/2 = 1.0512492×10⁻⁷ ⇒ pH 6.97829 (not 8)

常见问题

How do you calculate pH from concentration?

For a strong acid, [H⁺] equals the acid concentration times the protons each unit releases, and pH = −log₁₀[H⁺]: 0.01 M HCl has pH 2.00. For a strong base, take pOH = −log₁₀[OH⁻] and subtract it from 14: 0.005 M Ba(OH)₂ releases 0.01 M OH⁻, so pOH is 2 and pH is 12. Weak acids only partly ionise, so they also need Ka.

Why is the pH of pure water 7?

Water ionises slightly, H₂O ⇌ H⁺ + OH⁻, and at 25 °C the product [H⁺][OH⁻] equals Kw = 1.0 × 10⁻¹⁴. In pure water the two concentrations are equal, so each is 10⁻⁷ mol/L and the pH is 7. Kw grows with temperature: at 50 °C pKw is 13.26, so neutral water has pH 6.63. This calculator uses the 25 °C value.

What is the difference between Ka and pKa?

Ka is the acid dissociation constant, [H⁺][A⁻]/[HA] at equilibrium, and pKa = −log₁₀Ka. A lower pKa means a stronger acid, and each unit of pKa is a factor of 10 in Ka. Acetic acid has pKa 4.76, so Ka = 1.74 × 10⁻⁵ (often rounded to 1.8 × 10⁻⁵); hydrofluoric acid, Ka = 6.8 × 10⁻⁴, has pKa 3.17.

When is the Henderson–Hasselbalch equation accurate?

pH = pKa + log₁₀([A⁻]/[HA]) holds when the acid and its conjugate base are both far more concentrated than Ka and [H⁺], and their ratio lies between 0.1 and 10, which keeps the pH within one unit of pKa. Outside that ratio the buffer has little capacity left, and the calculator warns you. Equal concentrations give pH = pKa, so a 1:1 acetate buffer sits at 4.76.

Can pH be negative?

Yes. A 2 mol/L solution of a strong acid such as HCl has [H⁺] of about 2 mol/L, and −log₁₀2 gives pH −0.30. At concentrations like this, ion activities differ markedly from concentrations, so a pH electrode reads a different value. Treat any calculated pH below 0 or above 14 as an estimate rather than a measurement.

“pH calculator for acids, bases and buffers”有多准确?

准确性取决于输入值和方法的假设。十进制运算使用50位有效数字,但估算、数值方法和源数据的精度可能较低;显示时的舍入并不能消除这些限制。 已按独立来源核验的计算示例:8。 例如,“[H⁺] = 0.001 mol/L”根据−log₁₀(10⁻³) = 3; pOH = 14 − 3进行核验。

这种方法出自哪里?

OpenStax Chemistry 2e, §14.2 pH and pOH; §14.3 Relative strengths of acids and bases; §14.6 Buffers; IUPAC Gold Book — pH; ionic product of water (Kw = 1.0 × 10⁻¹⁴ at 25 °C).

关于此计算器

pH=−log⁡10[H+], pH+pOH=14;x2+Kax−KaC=0;pH=pKa+log⁡10[A−][HA]\text{pH} = -\log_{10}[\mathrm{H^+}],\ \text{pH} + \text{pOH} = 14;\quad x^2 + K_a x - K_aC = 0;\quad \text{pH} = \text{p}K_a + \log_{10}\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}

来源

  1. OpenStax Chemistry 2e, §14.2 pH and pOH; §14.3 Relative strengths of acids and bases; §14.6 Buffers
  2. IUPAC Gold Book — pH; ionic product of water (Kw = 1.0 × 10⁻¹⁴ at 25 °C)

已对照来源验证

此计算器包含 8 个已解示例,答案来自独立来源。这些示例会在测试套件中运行,你也可以在此运行验证。

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