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 × 10^ -5 , pK_ a =4.76)

Correct answer: D. 4.76 < pH < 5.0 aligned pH & =pK_a+ [ salt ] [ acid ] & =4.76+ ((7.5)/(500))/((5)/(500))=4.7+ 1.5=4.87 aligned Hence correct answer is 4.76 < pH < 5.0

JEE Main 201325 AprilOnline

Exam: JEE Main
Year: 2013
Date: 25 April
Mode: Online
Subject: Chemistry
Chapter: Ionic Equilibrium

JEE Main 2013 · 25 April · Online · Chemistry · Ionic Equilibrium

What would be the pH of a solution obtained by mixing 5 ~g of...

What would be the

\mathrm{pH}

of a solution obtained by mixing

5 \mathrm{~g}

of acetic acid and

7.5 \mathrm{~g}

of sodium acetate and making the volume equal to

500 \mathrm{~mL}

\left(\mathrm{K}_{\mathrm{a}}=1.75 \times 10^{-5}, \mathrm{pK}_{\mathrm{a}}=4.76\right)

Options

A $\mathrm{pH}=4.70$
B $\mathrm{pH} < 4.70$
C $\mathrm{pH}$ of solution will be equal to $\mathrm{pH}$ of acetic acid
D $4.76 < \mathrm{pH} < 5.0$

Correct answer

D. $4.76 < \mathrm{pH} < 5.0$

Step-by-step solution

\begin{aligned} \mathrm{pH} & =\mathrm{pK}_a+\log \frac{[\text{ salt }]}{[\text{ acid }]} \\ & =4.76+\log \frac{\frac{7.5}{500}}{\frac{5}{500}}=4.7+\log 1.5=4.87 \end{aligned}

Hence correct answer is

4.76 < \mathrm{pH} < 5.0

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