The pH of a buffer solution prepared by mixin | vedclass

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Question

(B) The $pH$ of an acidic buffer is given by the Henderson-Hasselbalch equation: $pH = pKa + \log \frac{[Salt]}{[Acid]}$.Given: $pH = 4.8$,$pKa = 4.8$

Vedclass · Accepted Answer

$0.04$

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(B) The $pH$ of an acidic buffer is given by the Henderson-Hasselbalch equation: $pH = pKa + \log \frac{[Salt]}{[Acid]}$.Given: $pH = 4.8$,$pKa = 4.8$,$V_{acid} = 50 \ mL$,$M_{acid} = 0.2 \ M$,$V_{salt} = 250 \ mL$.Substituting the values: $4.8 = 4.8 + \log \frac{[Salt]}{[Acid]}$.This implies $\log \frac{[Salt]}{[Acid]} = 0$,so $\frac{[Salt]}{[Acid]} = 10^0 = 1$.Therefore,$[Salt] = [Acid]$.Calculate the moles of acid: $n_{acid} = M 	imes V = 0.2 \ M 	imes 50 \ mL = 10 \ mmol$.Since $[Salt] = [Acid]$,the moles of salt must also be $10 \ mmol$ in the final mixture.Concentration of salt $[Salt] = \frac{n_{salt}}{V_{total}} = \frac{10 \ mmol}{(50 + 250) \ mL} = \frac{10}{300} \ M = 0.0333 \ M$.Wait,re-evaluating the question context: If the question implies the concentration of the stock solution of $CH_3COONa$ added,let $M_{salt}$ be the concentration. Then $n_{salt} = M_{salt} 	imes 250 \ mL$. Since $[Salt] = [Acid]$ in the mixture,$\frac{M_{salt} 	imes 250}{300} = \frac{0.2 	imes 50}{300}$.Thus,$M_{salt} 	imes 250 = 10$,so $M_{salt} = \frac{10}{250} = 0.04 \ M$.

The $pH$ of a buffer solution prepared by mixing $50 \ mL$ of $0.2 \ M$ $CH_3COOH$ and $250 \ mL$ of $CH_3COONa$ is $4.8$. What is the concentration of $CH_3COONa$ (in $M$)? Given that the $pKa$ of $CH_3COOH$ is $4.8$.

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