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(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

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$107$

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(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 $NH_4Cl = 	ext{moles} 	imes 	ext{molar mass} = 2 \, mol 	imes 53.5 \, g \, mol^{-1} = 107 \, g$.

Buffer Solution	Volume of $0.1 \ M$ Weak Acid (mL)	Volume of $0.1 \ M$ Sodium Salt (mL)
$I$	$4.0$	$4.0$
$II$	$4.0$	$40.0$
$III$	$40.0$	$4.0$
$IV$	$0.1$	$10.0$

Similar Questions

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 . . . . . . $\times 10^{-2}$.

Buffer Solution Volume of $0.1 \ M$ Weak Acid (mL) Volume of $0.1 \ M$ Sodium Salt (mL)

$I$ $4.0$ $4.0$

$II$ $4.0$ $40.0$

$III$ $40.0$ $4.0$

$IV$ $0.1$ $10.0$

Which of the two sets of buffer solutions have the least $pH$?

If $12.5 \ mL$ of $0.1 \ M$ $KOH$ is added into $50 \ mL$ of $0.1 \ M$ $CH_3COOH$ solution,then what would be the final $pH$? $(K_a(CH_3COOH) = 2 \times 10^{-5})$

When $0.02 \ mol$ of $NaOH$ is added to $1 \ L$ of a buffer solution,its $pH$ changes from $5.75$ to $5.80$. What is the buffer capacity of the solution?

What is the total number of buffers formed during the titration of $H_3PO_4$ with $NaOH$?

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