What would be the pH of a solution obtained b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The solution is a buffer solution consisting of a weak acid (acetic acid) and its salt (sodium acetate).The Henderson-Hasselbalch equation is: $pH

Vedclass · Accepted Answer

$4.76 < pH < 5.0$

Vedclass · Answer

(D) The solution is a buffer solution consisting of a weak acid (acetic acid) and its salt (sodium acetate).The Henderson-Hasselbalch equation is: $pH = pK_a + \log \frac{[salt]}{[acid]}$First,calculate the molar concentrations (or simply the ratio of moles since the volume is the same for both):Moles of acetic acid $(CH_3COOH)$ = $\frac{5 \ g}{60 \ g/mol} = 0.0833 \ mol$Moles of sodium acetate $(CH_3COONa)$ = $\frac{7.5 \ g}{82 \ g/mol} = 0.0915 \ mol$Using the ratio of moles in the Henderson-Hasselbalch equation:$pH = 4.76 + \log \left( \frac{0.0915}{0.0833} \right) = 4.76 + \log(1.098) \approx 4.76 + 0.04 = 4.80$Since $4.80$ falls in the range $4.76 < pH < 5.0$,the correct option is $D$.

What would be the $pH$ of a solution obtained by mixing $5 \ g$ of acetic acid and $7.5 \ g$ of sodium acetate and making the volume equal to $500 \ mL$? $(K_a = 1.75 \times 10^{-5}, pK_a = 4.76)$

Similar Questions

$0.1 \ mol$ of $CH_3NH_2$ $(K_b = 5 \times 10^{-4})$ is mixed with $0.08 \ mol$ of $HCl$ and diluted to $1 \ L$. The $[H^{+}]$ in the solution is:

$A$ solution of $H_3BO_3$ and borax is known as:

Which of the following buffer solutions will have the highest acidic character?

$0.1 \, M$ acetic acid solution is titrated against $0.1 \, M \, NaOH$ solution. What would be the difference in $pH$ between $1/4$ and $3/4$ stages of neutralization of acid?

$A$ buffer solution has equal volumes of $0.2 \ M \ NH_4OH$ and $0.02 \ M \ NH_4Cl$. The $pK_b$ of the base is $5$. The $pH$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module