The following table has some acids and their conjugate bases listed. Going down the column, the acid strength decreases while the base strength increases. We can conclude from this table that a stronger acid has a weaker conjugate base and a weaker acid has a stronger conjugate base.

A strong acid essentially completely dissociates in solution. The equation for a strong acid is

HA (aq) + H₂O (l) → H₃O⁺ (aq) + A^– (aq)

A weak acid only partially dissociates, and we use a double arrow to show the reaction reaches an equilibrium.

HA (aq) + H₂O (l) ⇄ H₃O⁺ (aq) + A^– (aq)

A weak acid has an equilibrium constant expression, K_a, called the acid dissociation constant or the acid-ionization constant.

HA (aq) + H₂O (l) ⇄ H₃O⁺ (aq) + A^– (aq)    \(\displaystyle K_a\;=\;\frac{[H_3O^+][A^-]}{[HA]}\)
 
Note, liquid water does not appear in the acid dissociation constant expression. The concentration of liquid water is 55.4 M.

In the table below are the relationship between Ka and pKa. The pK_a is equal to -log[Ka]. Note, a weaker acid will have a higher pK_a than a stronger acid.

An acid is weaker if it has a lower K_a value. For example, acetic acid has K_a equal to 1.8 x 10^-5 and hydrocyanic acid, HCN, has K_a equal to 4.9 x 10^-10. The HCN is the weaker acid because its K_a value is smaller than that of acetic acid. In the table below are acids with their K_a values and corresponding pK_a.

Below are some problem solving tips for weak acids. We see that the hydronium concentration from the self-ionization of water is negligible. In general, a weak acid has a small Ka, and its dissociation is also negligible, [HA]_eq ≅ [HA]_initial. Always check assumptions using the 5% rule.

If you were asked what is the pH for a solution that is 0.10 M HNO₃, your answer would be a pH of 1.00. But, what about the pH of a 0.10 M formic acid (HCOOH) solution? In this case we have a weak acid that only partially dissociates. We need the value of Ka. For HCOOH, Ka = 1.8 x 10^-4. First we write the balanced chemical equation and the Ka expression.

HCOOH (aq) + H₂O (l) ⇄ COOH^– (aq) + H₃O⁺ (aq)

\(\displaystyle Ka\;=\;\frac{[\mathrm{COOH^-}][\mathrm{H_3O^+}]}{\mathrm{HCOOH}}\;=\;1.8\times 10^{-4}\)

Next, we write our ICE table.

Assume that 0.10 M – x ≅ 0.10 M < 5% Now, put the equilibrium concentrations into the Ka expression.

\(\displaystyle \frac{[x][x]}{0.10}\;=\;1.8\times 10^{-4}\)

Solve for x.

\(\displaystyle x\;=\;\sqrt{Ka\times [HA]_{init}}\;=\;\sqrt{1.8\times 10^{-4}\times 0.10\;M\;=\;4.2\times 10^{-3}\;M}\)

Check the assumption.

\(\displaystyle \frac{x}{[\mathrm{HCOOH}]_{init}}\times 100\;=\;\frac{4.2\times 10^{-3}\;M}{0.10\;M}\times 100\;=\;4.2%\)

Our assumption is valid since 4.2% < 5%. Finally, we calculate the pH. Recall, x = [H₃O⁺] = 4.2 x 10^-3 M.

pH = -log[H₃O⁺] = -log(4.2 x 10^-3) = 2.38.

The pH is higher for the weaker acid, HCOOH, because it is only partially dissociated. The percent dissociation (ionization) is calculated as

\(\displaystyle \%Dissociation\;=\;\frac{x}{[\mathrm{HCOOH}]_{init}}\times 100\;=\;\frac{4.2\times 10^{-3}\;M}{0.10\;M}\times 100\;=\;4.2\%\)

The 0.10 M HCOOH is only 4.2% dissociated. It is important to always check the assumption using the 5% rule because x is not always negligible.

Watch the following video

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Exercises

Exercise 1. What is the pH of aqueous 0.25 M nitrous acid, HNO₂? Ka = 4.5 x 10^-4.

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Exercise 2. An aqueous hydrozoic acid, HN₃, solution has a pH of 3.26. What was the initial concentration of the solution? Ka = 1.9 x 10^-5.

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Exercise 3. The percent ionization of ascorbic acid is 0.35%. What was the initial concentration of the ascorbic acid if the pH of the solution is 4.85? Ka = 1.8 x 10^-5.

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Exercise 4. Calculate the pH of an aqueous solution that is 0.025 M in formic acid, HCOOH. Ka = 1.8 x 10^-4. What is the percent dissociation?

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Exercise 5. Calculate the pH, pOH, and the concentrations of all species in 0.45 M HF. Ka = 3.5 x 10^-4.

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Exercise 6. A benzoic acid, C₆H₅COOH, solution has a pH of 2.68. What is the percent dissociation of this solution? Ka = 6.5 x 10^-5

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Exercise 7. Vinegar contains 5.0% by mass of acetic acid (CH₃COOH) in water. If the pH of the solution is 2.50 and Ka = 1.8 x 10^-5, what is the percent dissociation of CH₃COOH in vinegar? The density of vinegar is 1.00 g/mL.

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