(D) For a basic buffer,$pOH = pK_b + \log \frac{[Salt]}{[Base]}$.Given $pH = 9$,so $pOH = 14 - 9 = 5$.$pK_b = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8

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(D) For a basic buffer,$pOH = pK_b + \log \frac{[Salt]}{[Base]}$.Given $pH = 9$,so $pOH = 14 - 9 = 5$.$pK_b = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$.Substituting the values: $5 = 4.745 + \log \frac{[NH_4Cl]}{[NH_4OH]}$.$0.255 = \log \frac{[NH_4Cl]}{1.0}$.$[NH_4Cl] = 10^{0.255} \approx 1.8 \, M$.Since the volume is $1.0 \, L$,the number of moles of $NH_4Cl$ is $1.8 \, mol$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Buffer solution

Medium

$A$ buffer solution with $pH = 9$ is prepared by mixing $NH_4Cl$ and $NH_4OH$. Calculate the number of moles of $NH_4Cl$ dissolved in $1.0 \, L$ of $1.0 \, M \, NH_4OH$ solution. (Given: $K_b(NH_4OH) = 1.8 \times 10^{-5}$)

A
$3.6$
B
$2.6$
C
$1.5$
D
$1.8$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Buffer solution

The $pH$ of a simple sodium acetate buffer is given by $pH = pK_a + \log \frac{[Salt]}{[Acid]}$. $K_a$ of acetic acid $= 1.8 \times 10^{-5}$. If $[Salt] = [Acid] = 0.1 \, M$,the $pH$ of the solution would be about

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$A$ buffer solution is prepared by mixing $10 \ mL$ of $1.0 \ M$ acetic acid and $20 \ mL$ of $0.5 \ M$ sodium acetate and then diluted to $100 \ mL$ with distilled water. The $pH$ of the buffer solution is ($pK_a$ of acetic acid is $4.76$).

MediumAP EAMCET 2019

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Which of the following combinations will form a basic buffer solution?

DifficultNEET 2019

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The $pH$ of a buffer solution decreases by $0.02$ units when $0.12 \ g$ of acetic acid is added to $250 \ mL$ of a buffer solution of acetic acid and potassium acetate at $27^{\circ} C$. The buffer capacity of the solution is

DifficultTS EAMCET 2009

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The $pH$ of a buffer solution made by mixing $25 \ mL$ of $0.02 \ M$ $NH_4OH$ and $25 \ mL$ of $0.2 \ M$ $NH_4Cl$ at $25^{\circ}C$ is ($pK_b$ of $NH_4OH = 4.8$).

DifficultAP EAMCET 2013

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$A$ buffer solution with $pH = 9$ is prepared by mixing $NH_4Cl$ and $NH_4OH$. Calculate the number of moles of $NH_4Cl$ dissolved in $1.0 \, L$ of $1.0 \, M \, NH_4OH$ solution. (Given: $K_b(NH_4OH) = 1.8 \times 10^{-5}$)

Similar Questions

The $pH$ of a simple sodium acetate buffer is given by $pH = pK_a + \log \frac{[Salt]}{[Acid]}$. $K_a$ of acetic acid $= 1.8 \times 10^{-5}$. If $[Salt] = [Acid] = 0.1 \, M$,the $pH$ of the solution would be about

$A$ buffer solution is prepared by mixing $10 \ mL$ of $1.0 \ M$ acetic acid and $20 \ mL$ of $0.5 \ M$ sodium acetate and then diluted to $100 \ mL$ with distilled water. The $pH$ of the buffer solution is ($pK_a$ of acetic acid is $4.76$).

Which of the following combinations will form a basic buffer solution?

The $pH$ of a buffer solution decreases by $0.02$ units when $0.12 \ g$ of acetic acid is added to $250 \ mL$ of a buffer solution of acetic acid and potassium acetate at $27^{\circ} C$. The buffer capacity of the solution is

The $pH$ of a buffer solution made by mixing $25 \ mL$ of $0.02 \ M$ $NH_4OH$ and $25 \ mL$ of $0.2 \ M$ $NH_4Cl$ at $25^{\circ}C$ is ($pK_b$ of $NH_4OH = 4.8$).

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