The pK_a of a weak acid is 5.85. The concentr | vedclass

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(B) The Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[	ext{conjugate base}]}{[	ext{acid}]}$.Given that the concentrations of

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$5$

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(B) The Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[	ext{conjugate base}]}{[	ext{acid}]}$.Given that the concentrations of the acid and its conjugate base are equal,we have $[	ext{conjugate base}] = [	ext{acid}]$,which implies $\frac{[	ext{conjugate base}]}{[	ext{acid}]} = 1$.Substituting the values into the equation: $pH = 5.85 + \log(1)$.Since $\log(1) = 0$,we get $pH = 5.85 + 0 = 5.85$.Therefore,the correct option is $B$.

The $pK_a$ of a weak acid is $5.85$. The concentrations of the acid and its conjugate base are equal at a $pH$ of (in $.85$)

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