I was solving a titration question and the reaction I came to was this:

$$ \begin{array}{lcccc} \ce{&BOH(aq) &+ &HCl(aq) -> &BC(aq) &+ &H2O(l) }\\ \text{Initial} & x & & & 0 & & 0 \\ \text{Final} & x/4 & & & 3x/4 & & 3x/4 \end{array} $$

Titration was done with $\ce{HCl}$ and $\ce{BOH}$ is a weak base. I got the answer treating this as a buffer but then I thought why would the salt formed i.e. $\ce{BCl}$ not get hydrolyzed and have an impact on the $\mathrm{pH}.$ Tried thinking about it and could only come to the conclusion that that $\ce{HA}$ being acid would have major contribution to the $\mathrm{pH}$ though I am not satisfied.

  • $\begingroup$ In a buffer, the concentration of both the weak base and the conjugate weak acid influence the pH. Could you edit your question to indicate states of matter and whether chloride is a spectator ion or covalently bound in the product? Then, it becomes possible to write the equilibrium constant expression and answer the question. $\endgroup$ – Karsten Theis Nov 28 '19 at 13:37
  • $\begingroup$ Oh I have written titration guess that implies that all states are liquid/aqueous and that thing related to chloride being a spectator ion was not mentioned in the question $\endgroup$ – naruto_022 Nov 28 '19 at 17:14
  • $\begingroup$ What is your question ? Are you trying to calculate the pH of the solution ? In this case, you simply apply the formula of a mixture acid + corresponding base. This formula takes care of the possible back-reaction. $\endgroup$ – Maurice Nov 28 '19 at 17:38
  • $\begingroup$ Yeah well the question was about finding the pH at 3/4th of the equvivalence point and I found the answer using the formula you just mentioned. My question is how does the formula take care that the salt formed is not hydrolyzed back. Is it something like because it is in equilibrium?? $\endgroup$ – naruto_022 Nov 29 '19 at 6:28

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