Assertion : In a titration of weak acid and N | vedclass

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(A) In the titration of a weak acid $(HA)$ with a strong base $(NaOH)$,the reaction is $HA + OH^- \rightarrow A^- + H_2O$. At the half equivalence poi

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(A) In the titration of a weak acid $(HA)$ with a strong base $(NaOH)$,the reaction is $HA + OH^- \rightarrow A^- + H_2O$. At the half equivalence point,half of the acid has been neutralized,meaning $[HA] = [A^-]$. According to the Henderson-Hasselbalch equation,$pH = pK_a + \log(\frac{[salt]}{[acid]})$. Since $[salt] = [acid]$ at this point,$\log(1) = 0$,so $pH = pK_a$. This mixture acts as an acidic buffer,and the buffer capacity is indeed maximum when the concentration of the acid equals the concentration of its conjugate base (salt).

Assertion : In a titration of weak acid and $NaOH$,the $pH$ at half equivalence point is $pK_a$.
Reason : At half equivalence point,it forms an acidic buffer and the buffer capacity is maximum where $[acid] = [salt]$.

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Assertion : In a titration of weak acid and $NaOH$,the $pH$ at half equivalence point is $pK_a$.Reason : At half equivalence point,it forms an acidic buffer and the buffer capacity is maximum where $[acid] = [salt]$.