Class XII students were asked to prepare one | vedclass

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

Question

(C) For a basic buffer,the $pH$ is related to $pOH$ by the equation: $pH + pOH = 14$. Given $pH = 8.26$,so $pOH = 14 - 8.26 = 5.74$. The Henderson-Has

Vedclass · Accepted Answer

$107$

Vedclass · Answer

(C) For a basic buffer,the $pH$ is related to $pOH$ by the equation: $pH + pOH = 14$. Given $pH = 8.26$,so $pOH = 14 - 8.26 = 5.74$. The Henderson-Hasselbalch equation for a basic buffer is: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$. Substituting the given values: $5.74 = 4.74 + \log \frac{[NH_4^+]}{0.2}$. $1 = \log \frac{[NH_4^+]}{0.2}$,which implies $\frac{[NH_4^+]}{0.2} = 10^1 = 10$. Therefore,$[NH_4^+] = 10 	imes 0.2 = 2 \, M$. Since the volume is $1 \, L$,the number of moles of $NH_4Cl$ required is $2 \, mol$. Mass of $NH_4Cl = 	ext{moles} 	imes 	ext{molar mass} = 2 \, mol 	imes 53.5 \, g \, mol^{-1} = 107 \, g$.

Similar Questions

Which of the following mixtures in water acts as a basic buffer?

Which of the following is a buffer solution?

The pH of a solution obtained by mixing $5 \text{ mL}$ of $0.1 \text{ M } NH_4OH$ solution with $250 \text{ mL}$ of $0.1 \text{ M } NH_4Cl$ solution is . . . . . . $\times 10^{-2}$.

$2 \ g$ acetic acid and $3 \ g$ sodium acetate are present in $100 \ mL$ aqueous solution. What will be the $pH$ of the solution if the ionisation constant of acetic acid is $1.8 \times 10^{-5}$?

The ratio of volumes of $CH_3COOH$ $0.1 \ N$ to $CH_3COONa$ $0.1 \ N$ required to prepare a buffer solution of $pH$ $5.74$ is (Given, $pK_a$ of $CH_3COOH$ is $4.74$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module