(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

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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

An example of a neutral buffer is:

$A$ buffer solution is prepared with a concentration of $0.30 \ M \ NH_3$ and $0.20 \ M \ NH_4^+$. If the equilibrium constant for $NH_3$ is $K_b = 1.8 \times 10^{-5}$,what will be the $pH$ of this solution?

$A$ weak acid of dissociation constant $10^{-5}$ is being titrated with aqueous $NaOH$ solution. The $pH$ at the point of one-third neutralization of the acid will be:-

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$A$ solution which is resistant to change of $pH$ upon the addition of an acid or a base is known as

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