Stoichiometric point of titration

Generally, the salt of a weak acid and a strong base (e.g. CH3COOH and NaOH) is a weak conjugate base, which hydrolyses in water, i.e. reacts with water to reform the acid according to the following reaction:

CH_3COO^-(aq)+H_2O(l)\; \begin{matrix}K_h\\ \rightleftharpoons \end{matrix}\; CH_3COOH(aq)+OH^-(aq)

where Kh is the hydrolysis constant. This results in a basic solution since OH is formed. During the stage of a strong base to weak acid titration between the start point and the stoichiometric point, the presence of unneutralised acid in the reaction flask causes the equilibrium of the hydrolysis reaction to shift to the left. However, when the stoichiometric point is reached, the weak acid is completely neutralised by the base and we can no longer ignore the effects of water on the pH of the solution. Since,

K_a=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}\; \; \; and\; \; \; K_w=[H^+][OH^-]


From the hydrolysis equation, [CH3COOH] = [OH], so




Taking the logarithm on both sides of the above equation and applying the definitions of pH, pKa and pKw,

pH=\frac{pK_w+pK_a+log[A^-]}{2}\; \; \; \; \; \; \; \; (4)

where [A] = [CH3COO], the concentration of the salt at the stoichiometric point before hydrolysis. Eq4 is the general equation to calculate the pH of a strong base to weak acid titration at the stoichiometric point. 


Calculate the pH of the titration of 10 cm3 of 0.200 M of CH3COOH (Ka = 1.75 x 10-5) with 0.100 M of NaOH at the stoichiometric point.


Using eq4,

pH=\frac{-log10^{-14}-log\left ( 1.75\times 10^{-5}\right)+log\left ( \frac{0.100\times 0.02}{0.01+0.02} \right )}{2}=8.79

Note that [A] can be computed using either the acid or the base.


The same logic applies when determining the equation for the pH of a strong acid to weak base titration at the stoichiometric point, with the expected pH lower than 7.


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