A buffer solution contains 100 mL of 0.01 M | vedclass

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(D) The $pH$ of a buffer solution is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log_{10} \frac{[Salt]}{[Acid]}$.Initial $pH$ c

Vedclass · Accepted Answer

$5.34, 5.34$

Vedclass · Answer

(D) The $pH$ of a buffer solution is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log_{10} \frac{[Salt]}{[Acid]}$.Initial $pH$ calculation:$pH = 4.74 + \log \frac{(0.02 	imes 200) / 300}{(0.01 	imes 100) / 300} = 4.74 + \log \frac{4}{1} = 4.74 + 0.60 = 5.34$.After adding $700 \ mL$ of $1 \ M \ NaCl$:$NaCl$ is a neutral salt and does not participate in the acid-base equilibrium. Adding it only dilutes the solution.Since the ratio of the concentration of salt to acid remains unchanged upon dilution,the $pH$ remains constant at $5.34$.

$A$ buffer solution contains $100 \ mL$ of $0.01 \ M \ CH_3COOH$ and $200 \ mL$ of $0.02 \ M \ CH_3COONa$. If $700 \ mL$ of $1 \ M \ NaCl$ is added,what are the $pH$ values before and after the addition? $(pK_a = 4.74)$

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$A$ buffer solution of $pH = 9.80$ is to be prepared from $1 \ M \ NH_4Cl$ and $1 \ M \ NH_3$ solution. $pK_{b}$ of $NH_3$ is $4.74$. In $1 \ L$ of the buffer solution,the volume of $NH_4Cl$ solution required is $(\log 0.28 = -0.54) :-$ (in $mL$)

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