The OH^{-} concentration in a mixture of 5.0 | vedclass

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(A) The mixture consists of a weak base $(NH_{3})$ and its salt $(NH_{4}Cl)$,forming a basic buffer.First,calculate the moles of each component:$n(NH_

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$3$

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(A) The mixture consists of a weak base $(NH_{3})$ and its salt $(NH_{4}Cl)$,forming a basic buffer.First,calculate the moles of each component:$n(NH_{4}Cl) = 0.0504 \ M 	imes 5.0 \ mL = 0.252 \ mmol$$n(NH_{3}) = 0.0210 \ M 	imes 2.0 \ mL = 0.042 \ mmol$Using the Henderson-Hasselbalch equation for a basic buffer:$[OH^{-}] = K_{b} 	imes \frac{[Base]}{[Salt]} = K_{b} 	imes \frac{n(NH_{3})}{n(NH_{4}Cl)}$$[OH^{-}] = 1.8 	imes 10^{-5} 	imes \frac{0.042}{0.252}$$[OH^{-}] = 1.8 	imes 10^{-5} 	imes \frac{1}{6} = 0.3 	imes 10^{-5} = 3 	imes 10^{-6} \ M$Comparing this with $x 	imes 10^{-6} \ M$,we get $x = 3$.

The $OH^{-}$ concentration in a mixture of $5.0 \ mL$ of $0.0504 \ M \ NH_{4}Cl$ and $2 \ mL$ of $0.0210 \ M \ NH_{3}$ solution is $x \times 10^{-6} \ M$. The value of $x$ is ..... .
(Nearest integer)
$[ \text{Given } K_{w}=1 \times 10^{-14} \text{ and } K_{b}=1.8 \times 10^{-5} ]$

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The $OH^{-}$ concentration in a mixture of $5.0 \ mL$ of $0.0504 \ M \ NH_{4}Cl$ and $2 \ mL$ of $0.0210 \ M \ NH_{3}$ solution is $x \times 10^{-6} \ M$. The value of $x$ is ..... .(Nearest integer)$[ \text{Given } K_{w}=1 \times 10^{-14} \text{ and } K_{b}=1.8 \times 10^{-5} ]$