Start point of titration

We shall make two assumptions:

    • [H+] from water is negligible, i.e. H+ in the flask containing the weak acid is solely due to that of the acid, [H+]a, as the dissociation of water is suppressed at this stage.
    • For a weak acid, [HA]  is approximately equal to the concentration of the undissociated acid, [HA]ud, i.e. the dissociation of the weak acid HA is negligible.

The equation for the dissociation of a weak monoprotic acid is:

HA(aq)\rightleftharpoons H^+(aq)+A^-(aq)

with the equilibrium constant at the start of the titration as:

K_a=\frac{[H^+][A^-]}{[HA]}\approx \frac{[H^+]_a[A^-]}{[HA]_{ud}}=\frac{[H^+]_a\, ^2}{[HA]_{ud}}

Taking the logarithm on both sides of the above equation and rearranging, we have:

pH=\frac{pK_a-log[HA]_{ud}}{2}\; \; \; \; \; \; \; \; (1)

Eq1 is the general formula for determining the pH of a strong base to weak acid titration at the start point.

The second assumption becomes less valid when the weak monoprotic acid or weak monoprotic base has Ka 10-3 or Kb 10-3 respectively. Removing the second assumption, the equilibrium constant expression becomes:

K_a\approx \frac{[H^+]_a[A^-]}{[HA]_{ud}-[H^+]_a}=\frac{[H^+]_a\, ^2}{[HA]_{ud}-[H^+]_a}

The corresponding pH equation is obtained by rearranging the above equation into a quadratic equation in terms of  [H+]a, finding the latter’s roots and taking the logarithm of the root:

pH=-log\left ( \frac{\sqrt{K_a\, ^2+4K_a[HA]_{ud}}-K_a}{2} \right )



Do both the above equation and eq1 give the same pH value for the titration of 10 cm3 of 0.200 M of CH3COOH (Ka = 1.75 x 10-5) with NaOH?


Yes, both formulae give pH = 2.73. This validates the applicability of eq1 for acids with Ka in the region of 10-5.


If we disregard both assumptions, we will end up deriving the complete pH titration curve for a strong base to weak acid titration. See this article in the advanced section for details.



Show that pKa + pKb = 14.


HA(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+A^-(aq)

A^-(aq)+H_2O(l)\rightleftharpoons HA(aq)+OH^-(aq)


K_a=\frac{[H_3O^+][A^-]}{[HA]}\; \; \; \; \; \; \; \; K_b=\frac{[HA][OH^-]}{[A^-]}




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