The pH of a buffer solution decreases by 0.02 | vedclass

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(D) Buffer capacity,$\beta = \frac{d C_{HA}}{d_{pH}}$,where $d C_{HA}$ is the number of moles of acid added per liter and $d_{pH}$ is the change in $p

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$0.4$

Vedclass · Answer

(D) Buffer capacity,$\beta = \frac{d C_{HA}}{d_{pH}}$,where $d C_{HA}$ is the number of moles of acid added per liter and $d_{pH}$ is the change in $pH$. First,calculate the number of moles of acetic acid added: $n = \frac{0.12 \ g}{60 \ g/mol} = 0.002 \ mol$. Next,calculate the concentration of acid added per liter $(d C_{HA})$: $d C_{HA} = \frac{0.002 \ mol}{0.250 \ L} = 0.008 \ mol/L$. Given the change in $pH$ $(d_{pH})$ is $0.02$. Therefore,$\beta = \frac{0.008}{0.02} = 0.4$.

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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