## Weak acids & bases

Strong acids ionize quantitatively, so we use a normal, one-directional arrow. Weak acids do not, so we have to use an equilibrium arrow. Acetic acid is an example of a weak acid:

CH3COOH(aq) ⇌ H+(aq) + CH3COO(aq).

Since the equilibrium position is not the same for all weak acids, we have to care about the equilibrium constant, which in this case we call the acid ionization constant. For the general reaction HA(aq) ⇌ H+(aq) + A(aq), the acid ionization constant is

K_"a" = (["H"^+]\["A"^-])/(["HA"]).

Similarly, weak bases have B(aq) + H2O(l) ⇌ HB+(aq) + OH(aq) as their general reaction. The base ionization constant is

K_"b" = (["HB"^+]\["OH"^-])/(["B"]).

Another way of representing the position of the equilibrium for weak acids is percent ionization:

p = (["H"^+])/(["HA"]) xx 100% qquad <=> qquad ["H"^+] = ["HA"]\(p/(100%)).

The following equation is true for all conjugate acid–base pairs:

K_"a"K_"b" = K_"w".

### Example

Given a solution of 0.082 mol/L phosphoric acid (a weak acid) whose pH is 1.68 upon reaching equilibrium, find the hydrogen ion concentration, the percent ionization, and the value of the acid ionization constant.

Finding the hydrogen ion concentration is easy enough:

["H"^+] = 10^(-"pH")\ "mol/L" = 10^-1.68\ "mol/L" = 0.021\ "mol/L".

We can use this to find the percent ionization:

p = (["H"^+])/(["HA"]) xx 100% = (0.02089\ "mol/L")/(0.082\ "mol/L") xx 100% = 26%.

Our dissociation reaction is H3PO4(aq) ⇌ H2PO4(aq) + H+(aq). We need to use an ICE table to find the equilibrium concentrations.

 H3PO4(aq) H2PO4−(aq) H+(aq) I 0.082 0 0 C −0.021 +0.021 +0.021 E 0.061 0.021 0.021

Now we can calculate the acid ionization constant:

K_"a" = (["H"^+]\["A"^-])/(["HA"]) = ((0.02089\ "mol/L")(0.02089\ "mol/L"))/(0.06111\ "mol/L") = 7.1xx10^-3.