Start point of titration

The start point of a titration marks the initial pH of the solution before any titrant is added, serving as a crucial reference for analysing the subsequent changes in acidity or alkalinity throughout the process.

To characterised the start point, we begin with the following 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 K_a≥ 10^-3 or K_b≥ 10^-3respectively. 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:

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

Question

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

Answer

Yes, both formulae give pH = 2.73. This validates the applicability of eq1 for acids with K_a 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.

Question

Show that pK_a + pK_b= 14.

Answer

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

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

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

$-logK_a-logK_b=-log\frac{[H_3O^+][A^-]}{[HA]}-log\frac{[HA][OH^-]}{[A^-]}$

$pK_a+pK_b=-log[H_3O^+][OH^-]=pK_w=14$

Start point of titration

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