Henderson's equation is pH = pK_a + logfrac{[ | vedclass

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(A) The Henderson-Hasselbalch equation is given by: $pH = pK_a + \log\frac{[	ext{salt}]}{[	ext{acid}]}$.When the acid is half-neutralized,the concen

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$4.3$

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(A) The Henderson-Hasselbalch equation is given by: $pH = pK_a + \log\frac{[	ext{salt}]}{[	ext{acid}]}$.When the acid is half-neutralized,the concentration of the salt formed is equal to the concentration of the remaining acid,i.e.,$[	ext{salt}] = [	ext{acid}]$.Substituting this into the equation: $pH = pK_a + \log(1)$.Since $\log(1) = 0$,we get $pH = pK_a$.Given $pK_a = 4.30$,therefore $pH = 4.30$.

Henderson's equation is $pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]}$. If the acid is half-neutralized,the value of $pH$ will be: $[pK_a = 4.30]$

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