pH of a mixture of 1 M HA (pK_a = 4.2) and | vedclass

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(C) The Henderson-Hasselbalch equation for an acidic buffer is given by:$pH = pK_a + \log \frac{[Salt]}{[Acid]}$Given that the concentration of both $

Vedclass · Accepted Answer

$100$

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(C) The Henderson-Hasselbalch equation for an acidic buffer is given by:$pH = pK_a + \log \frac{[Salt]}{[Acid]}$Given that the concentration of both $HA$ and $NaA$ is $1 \ M$,the ratio of their concentrations is equal to the ratio of their volumes.Let the volume of $HA$ be $x \ mL$. Then the volume of $NaA$ is $(300 - x) \ mL$.Substituting the values into the equation:$4.5 = 4.2 + \log \frac{300 - x}{x}$$0.3 = \log \frac{300 - x}{x}$Since $\log 2 = 0.3$,we have:$\log 2 = \log \frac{300 - x}{x}$$2 = \frac{300 - x}{x}$$2x = 300 - x$$3x = 300$$x = 100 \ mL$

$pH$ of a mixture of $1 \ M \ HA$ $(pK_a = 4.2)$ and $1 \ M \ NaA$ is $4.5$. In $300 \ mL$ of buffer,the volume of $HA$ is $....... \ mL$ $[\log \ 2 = 0.3]$

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