40 mL of 0.1 M ammonia solution is mixed wi | vedclass

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(C) The reaction is: $NH_3 + HCl 	o NH_4Cl$. Initial moles of $NH_3 = 40 \ mL 	imes 0.1 \ M = 4 \ mmol$. Initial moles of $HCl = 20 \ mL 	imes 0.1

Vedclass · Accepted Answer

$9.26$

Vedclass · Answer

(C) The reaction is: $NH_3 + HCl 	o NH_4Cl$. Initial moles of $NH_3 = 40 \ mL 	imes 0.1 \ M = 4 \ mmol$. Initial moles of $HCl = 20 \ mL 	imes 0.1 \ M = 2 \ mmol$. After the reaction,$2 \ mmol$ of $NH_3$ remains and $2 \ mmol$ of $NH_4Cl$ is formed. This forms a basic buffer solution. Using the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$. Since the volume is the same for both,we can use the ratio of moles: $pOH = 4.74 + \log \frac{2}{2} = 4.74 + 0 = 4.74$. Finally,$pH = 14 - pOH = 14 - 4.74 = 9.26$.

$40 \ mL$ of $0.1 \ M$ ammonia solution is mixed with $20 \ mL$ of $0.1 \ M \ HCl$. What is the $pH$ of the mixture? ($pK_b$ of ammonia solution is $4.74$).

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