A buffer solution contains 0.1 mol of sodium | vedclass

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(D) The $pH$ of a buffer solution is given by the Henderson-Hasselbalch equation: $pH = p K_{a} + \log \frac{[salt]}{[acid]}$.Initially,the solution c

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$p K_{a} + \log 2$

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(D) The $pH$ of a buffer solution is given by the Henderson-Hasselbalch equation: $pH = p K_{a} + \log \frac{[salt]}{[acid]}$.Initially,the solution contains $0.1 \ mol$ of sodium acetate in $1000 \ cm^{3}$ $(1 \ L)$ of $0.1 \ M$ acetic acid.After adding an additional $0.1 \ mol$ of sodium acetate,the total amount of salt becomes $0.1 + 0.1 = 0.2 \ mol$.The concentration of salt $[salt] = \frac{0.2 \ mol}{1 \ L} = 0.2 \ M$.The concentration of acid $[acid] = 0.1 \ M$.Substituting these values into the equation: $pH = p K_{a} + \log \frac{0.2}{0.1}$.$pH = p K_{a} + \log 2$.

$A$ buffer solution contains $0.1 \ mol$ of sodium acetate dissolved in $1000 \ cm^{3}$ of $0.1 \ M$ acetic acid. To the above buffer solution,$0.1 \ mol$ of sodium acetate is further added and dissolved. The $pH$ of the resulting buffer is

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Aqueous solutions of $HNO_3$,$KOH$,$CH_3COOH$,and $CH_3COONa$ of identical concentrations are provided.
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The $pH$ of a buffer solution made by mixing $25 \ mL$ of $0.02 \ M$ $NH_4OH$ and $25 \ mL$ of $0.2 \ M$ $NH_4Cl$ at $25^{\circ}C$ is ($pK_b$ of $NH_4OH = 4.8$).

The acidic and basic strength of buffer solutions remains constant because ...

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