(D) Buffer capacity, $\beta = \frac{dC_{HA}}{dpH}$.Here, $dC_{HA}$ is the number of moles of acid added per liter of the solution.$dC_{HA} = \frac{\te

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(D) Buffer capacity, $\beta = \frac{dC_{HA}}{dpH}$.Here, $dC_{HA}$ is the number of moles of acid added per liter of the solution.$dC_{HA} = \frac{\text{moles of acetic acid}}{\text{volume in liters}} = \frac{0.12 / 60}{250 / 1000} = \frac{0.002}{0.25} = 0.008 \ M$.The change in $pH$ is $dpH = 0.02$.Therefore, $\beta = \frac{0.008}{0.02} = 0.4$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Buffer solution

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The $pH$ of a buffer solution decreases by $0.02$ units when $0.12 \ g$ of acetic acid is added to $250 \ mL$ of a buffer solution of acetic acid and potassium acetate at $27^{\circ} C$. The buffer capacity of the solution is

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$0.1$
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$10$
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$1$
D
$0.4$

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Which of the following is not a buffer solution?

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The $pH$ of a buffer solution decreases by $0.02$ units when $0.12 \ g$ of acetic acid is added to $250 \ mL$ of a buffer solution of acetic acid and potassium acetate at $27^{\circ} C$. The buffer capacity of the solution is

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For a weak acid $(HA)$,$pK_a = 4.5$. The $pOH$ of a buffer solution of $HA$ in which $50\%$ of the acid is ionized is:

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