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Jul23-06, 07:13 AM
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6. pH of Weak Bases
Weak bases do not fully dissociate when placed in solution, an equilibrium is obtained similar to weak acids;

[tex]BOH(aq) \rightleftharpoons B^{+}(aq) + OH^{-}(aq)[/tex]

As with the acidic equilibrium, we can defined an equilibrium constant, which in this case is known as the basicity constant (Kb);

[tex]K_{b} = \frac{\left[ B^{+}(aq)\right] \left[ OH^{-}(aq) \right]}{\left[ BOH(aq) \right]}[/tex]

Again, here we can make a number of assumptions which greatly simplify the calculations. These assumptions hold when the base is not extremely weak or very dilute (i.e. we can ignore the ionic product of water). These assumptions are as follows and are similar to the assumptions made when calculating the pH of weak acids;
Assumption One: [B+(aq)] = OH-(aq)]
This means that we are ignoring the effect of the ionisation of water, this is a valid assumption for most 'normal' concentrations.
Assumption Two: As weak bases dissociate very little, we can assume that the concentration of the base at equilibrium is the same as if the base did not dissociate at all.
These assumptions allows us to rewrite the basicity constant as follows;

[tex]K_{b} = \frac{\left[ OH^{-}(aq) \right]^{2}}{\left[ BOH(aq) \right]}[/tex]

An expression for pH can then be found thus;

[tex]K_{b} = \frac{\left[ OH^{-}(aq) \right]^{2}}{\left[ BOH(aq) \right]} \Leftrightarrow \left[ OH^{-}(aq) \right] = \sqrt{K_{b} \cdot \left[ BOH(aq) \right]}[/tex]

Using the expression for the ionic product of water;

[tex]K_{w} = \left[ H^{+}(aq) \right] \left[OH^{-}(aq) \right] \Leftrightarrow \left[OH^{-}(aq) \right] = \frac{K_{w}}{\left[ H^{+}(aq) \right]}[/tex]

[tex]\frac{K_{w}}{\left[ H^{+}(aq) \right]} = \sqrt{K_{b} \cdot \left[ BOH(aq) \right]}[/tex]

[tex]\left[ H^{+}(aq) \right] = \frac{K_{w}}{\sqrt{K_{b} \cdot \left[ BOH(aq) \right]}}[/tex]

[tex]pH = -\log\left( \frac{K_{w}}{\sqrt{K_{b} \cdot \left[ BOH(aq) \right]}} \right)[/tex]

Using the laws of logarithms and knowing that pKw = 14 we can show that;

[tex]pH = 14 + \log\left( \sqrt{K_{b} \cdot \left[ BOH(aq) \right]}\right)[/tex]

Alternatively, one could take the equation for the ionisation product of water and find in the pH from the pOH of the basic solution;

[tex]\left[ OH^{-}(aq) \right] = \sqrt{K_{b} \cdot \left[ BOH(aq) \right]} \Leftrightarrow pOH = -\log\sqrt{K_{b} \cdot \left[ BOH(aq) \right]}[/tex]

[tex]K_{w} = \left[ H^{+}(aq) \right] \left[OH^{-}(aq) \right] \Leftrightarrow -\log K_{w} = pH \times pOH[/tex]

Using the laws of logarithms;

[tex]pH = -\log (1\times10^{-14}) - pOH[/tex]

[tex]pH = 14 + \log\left( \sqrt{K_{b} \cdot \left[ BOH(aq) \right]}\right)[/tex]