AI Math
Abrir app

Acid Dissociation Constant Calculator

Find Ka from pH or percent ionisation and convert between Ka, Kb, pKa and pKb, step by step
Ka of a 0.200 M acid that is 1.50% ionised
Kb of F- given Ka(HF) = 6.8 x 10^-4
Ka and pKa of a 0.0500 M acid with pH 3.20
Convert pKa = 3.17 to Ka

What the Dissociation Constant Measures

The acid dissociation constant is the equilibrium constant for an acid giving up a proton to water in dilute aqueous solution:

HA+H2O⇌H3O++A−,Ka=[H3O+][A−][HA]\mathrm{HA} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{A^-}, \qquad K_a = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}

All three concentrations are equilibrium values in mol/L. The solvent is omitted from the expression because it is present in enormous excess and its concentration is effectively constant.

KaK_a answers one question: of the acid molecules dissolved, what fraction have handed over a proton at equilibrium? A KaK_a near 10−210^{-2} describes an acid that is substantially ionised; 10−1010^{-10} describes one that is barely ionised at all.

The conjugate base. Every acid's KaK_a has a partner constant for its conjugate base, and at 25 °C the two are locked together by the ion product of water:

KaKb=Kw=1.0×10−14⟹pKa+pKb=14.00K_a K_b = K_w = 1.0 \times 10^{-14} \qquad\Longrightarrow\qquad \mathrm{p}K_a + \mathrm{p}K_b = 14.00

So a stronger acid necessarily has a weaker conjugate base. Both relations assume 25 °C, dilute solution, and that concentration is a good stand-in for activity.

Three Ways to Find Ka

From a measured pH

For a monoprotic weak acid at formal concentration CaC_a, the ionisation produces equal amounts of H3O+\mathrm{H_3O^+} and A−\mathrm{A^-}:

  1. x=[H3O+]=10−pHx = [\mathrm{H_3O^+}] = 10^{-\mathrm{pH}}, and [A−]=x[\mathrm{A^-}] = x as well.
  2. [HA]=Ca−x[\mathrm{HA}] = C_a - x — subtract what ionised.
  3. Ka=x2Ca−xK_a = \dfrac{x^2}{C_a - x}.

From percent ionisation

Percent ionisation is α=x/Ca×100%\alpha = x/C_a \times 100\%, so x=αCa/100x = \alpha C_a / 100. Substituting gives

Ka=Caα21−αK_a = \frac{C_a \alpha^2}{1 - \alpha}

with α\alpha as a decimal fraction. Note that percent ionisation rises on dilution even though KaK_a does not move.

From pKa

Ka=10−pKaK_a = 10^{-\mathrm{p}K_a}

Converting to Kb

Kb=KwKa=1.0×10−14KaK_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{K_a}

Significant figures

A pH measured to 2 decimal places gives [H3O+][\mathrm{H_3O^+}] to 2 significant figures, and therefore KaK_a to 2. Because KwK_w is quoted to 2 figures, any KbK_b derived from it inherits that limit.

Common Mistakes to Avoid

  • Using the formal concentration as [HA][\mathrm{HA}]. The denominator is Ca−xC_a - x; skipping the subtraction inflates KaK_a, badly so when ionisation exceeds a few percent.
  • Thinking KaK_a changes with concentration. It does not — only temperature moves it. Percent ionisation and pH do change with dilution.
  • Reporting KbK_b as 14−Ka14 - K_a. The additive relation is between the p-values; the constants themselves multiply to KwK_w.
  • Quoting a KaK_a for a strong acid. HCl and HNO3\mathrm{HNO_3} ionise essentially completely, so their equilibrium constants are not measurable by these methods in water.
  • Ignoring the second ionisation of a polyprotic acid. H2SO4\mathrm{H_2SO_4} and H3PO4\mathrm{H_3PO_4} have separate Ka1,Ka2,…K_{a1}, K_{a2}, \ldots, each far smaller than the last.
  • Assuming pKa+pKb=14\mathrm{p}K_a + \mathrm{p}K_b = 14 at every temperature — that sum is pKw\mathrm{p}K_w, which equals 14.00 only at 25 °C.

Examples

Step 1: Ionised amount: x=0.0150×0.200=3.00×10−3x = 0.0150 \times 0.200 = 3.00 \times 10^{-3} M, which equals both [H3O+][\mathrm{H_3O^+}] and [A−][\mathrm{A^-}]
Step 2: Remaining acid: [HA]=0.200−0.00300=0.19700[\mathrm{HA}] = 0.200 - 0.00300 = 0.19700 M
Step 3: Ka=(3.00×10−3)20.19700=9.00×10−60.19700K_a = \dfrac{(3.00 \times 10^{-3})^2}{0.19700} = \dfrac{9.00 \times 10^{-6}}{0.19700}
Step 4: Ka=4.5685×10−5K_a = 4.5685 \times 10^{-5}, kept to the 3 significant figures of the data
Answer: Ka=4.57×10−5K_a = 4.57 \times 10^{-5}

Step 1: pKa=−log⁡10(6.8×10−4)=4−log⁡10(6.8)=4−0.8325=3.1675\mathrm{p}K_a = -\log_{10}(6.8 \times 10^{-4}) = 4 - \log_{10}(6.8) = 4 - 0.8325 = 3.1675
Step 2: Kb=KwKa=1.0×10−146.8×10−4=1.4706×10−11K_b = \dfrac{K_w}{K_a} = \dfrac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} = 1.4706 \times 10^{-11}
Step 3: Both KwK_w and KaK_a carry 2 significant figures, so Kb=1.5×10−11K_b = 1.5 \times 10^{-11}
Step 4: pKb=14.00−3.17=10.83\mathrm{p}K_b = 14.00 - 3.17 = 10.83, which matches −log⁡10(1.47×10−11)=10.83-\log_{10}(1.47 \times 10^{-11}) = 10.83
Answer: pKa=3.17\mathrm{p}K_a = 3.17, Kb=1.5×10−11K_b = 1.5 \times 10^{-11}, pKb=10.83\mathrm{p}K_b = 10.83

Step 1: x=10−3.20=6.3096×10−4x = 10^{-3.20} = 6.3096 \times 10^{-4} M
Step 2: [HA]=0.0500−6.3096×10−4=0.049369[\mathrm{HA}] = 0.0500 - 6.3096 \times 10^{-4} = 0.049369 M
Step 3: Ka=(6.3096×10−4)20.049369=3.9811×10−70.049369=8.064×10−6K_a = \dfrac{(6.3096 \times 10^{-4})^2}{0.049369} = \dfrac{3.9811 \times 10^{-7}}{0.049369} = 8.064 \times 10^{-6}
Step 4: The pH has 2 decimal places, so KaK_a is reported to 2 significant figures
Step 5: pKa=−log⁡10(8.064×10−6)=6−0.9065=5.0935\mathrm{p}K_a = -\log_{10}(8.064 \times 10^{-6}) = 6 - 0.9065 = 5.0935
Step 6: Ionisation is 6.3096×10−4/0.0500=1.26%6.3096 \times 10^{-4}/0.0500 = 1.26\%, so the acid is genuinely weak here
Answer: Ka=8.1×10−6K_a = 8.1 \times 10^{-6}, pKa=5.09\mathrm{p}K_a = 5.09

Frequently Asked Questions

Put the equilibrium concentrations into Ka = [H3O+][A-]/[HA]. From a pH, x = 10^(-pH) gives both [H3O+] and [A-], and [HA] is the formal concentration minus x, so Ka = x²/(Ca - x). From percent ionisation, x is that fraction of Ca.

Ka = 10^(-pKa). A pKa of 3.17 gives Ka = 10^-3.17 = 6.8 x 10^-4. The conversion is exact; only the digits after the decimal point of the pKa are significant, so two decimals give two significant figures in Ka.

For a conjugate acid-base pair in water at 25 °C, Ka x Kb = Kw = 1.0 x 10^-14, so Kb = Kw/Ka. In log form, pKa + pKb = 14.00. Note that the constants multiply while the p-values add.

No. Ka is a true equilibrium constant and depends only on the acid and the temperature. Dilution raises the percent ionisation and raises the pH, but the value of Ka stays put — which is exactly why it is useful for comparing acids.

Related Solvers

Calculadora de equações quadráticas
Try AI-Math for Free

Get step-by-step solutions to any math problem. Upload a photo or type your question.

Start Solving