pH of a mixture of 1 M benzoic acid (pK_a = | vedclass

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(C) The Henderson-Hasselbalch equation for an acidic buffer is $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.Given: $pH = 4.5$,$pK_a = 4.2$,$[Salt] = 1 \ M

Vedclass · Accepted Answer

$100$

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(C) The Henderson-Hasselbalch equation for an acidic buffer is $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.Given: $pH = 4.5$,$pK_a = 4.2$,$[Salt] = 1 \ M$,$[Acid] = 1 \ M$.Let the volume of benzoic acid be $V \ mL$. Then the volume of $C_6H_5COONa$ is $(300 - V) \ mL$.Since the molarities are equal,the ratio of concentrations is equal to the ratio of volumes: $\frac{[Salt]}{[Acid]} = \frac{300 - V}{V}$.Substituting the values: $4.5 = 4.2 + \log \frac{300 - V}{V}$.$0.3 = \log \frac{300 - V}{V}$.Since $\log 2 = 0.3$,we have $\log 2 = \log \frac{300 - V}{V}$.$2 = \frac{300 - V}{V}$.$2V = 300 - V$.$3V = 300$.$V = 100 \ mL$.

$pH$ of a mixture of $1 \ M$ benzoic acid $(pK_a = 4.20)$ and $1 \ M$ $C_6H_5COONa$ is $4.5$. What is the volume of benzoic acid required to prepare a $300 \ mL$ buffer solution (in $mL$)? $[\log 2 = 0.3]$

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