Find pH of the resultant solution formed by t | vedclass

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(A) $1$. Calculate the moles of reactants: $n(Ba(OH)_2) = 0.5 \ L 	imes 0.1 \ M = 0.05 \ mol$. $n(OH^-) = 2 	imes 0.05 = 0.1 \ mol$. $n(NH_4Cl) = 0.

Vedclass · Accepted Answer

$8.7$

Vedclass · Answer

(A) $1$. Calculate the moles of reactants: $n(Ba(OH)_2) = 0.5 \ L 	imes 0.1 \ M = 0.05 \ mol$. $n(OH^-) = 2 	imes 0.05 = 0.1 \ mol$. $n(NH_4Cl) = 0.5 \ L 	imes 0.6 \ M = 0.3 \ mol$. $2$. Reaction: $Ba(OH)_2 + 2NH_4Cl ightarrow BaCl_2 + 2NH_3 + 2H_2O$. $0.1 \ mol$ of $OH^-$ reacts with $0.1 \ mol$ of $NH_4^+$ to form $0.1 \ mol$ of $NH_3$. Remaining $NH_4^+ = 0.3 - 0.1 = 0.2 \ mol$. $3$. The solution contains $NH_3$ $(0.1 \ mol)$ and $NH_4^+$ $(0.2 \ mol)$,forming a basic buffer. Total volume = $0.5 + 0.5 + 1.0 = 2.0 \ L$. $[NH_3] = 0.1 / 2 = 0.05 \ M$,$[NH_4^+] = 0.2 / 2 = 0.1 \ M$. $4$. Using Henderson-Hasselbalch equation: $pOH = pK_b + \log([NH_4^+] / [NH_3]) = -\log(10^{-5}) + \log(0.1 / 0.05) = 5 + \log(2) = 5 + 0.3 = 5.3$. $pH = 14 - pOH = 14 - 5.3 = 8.7$.

Find $pH$ of the resultant solution formed by the addition of $500 \ mL$ $0.1 \ M$ $Ba(OH)_2$ and $500 \ mL$ $0.6 \ M$ $NH_4Cl$ in $1 \ L$ pure $H_2O$ at $25 \ ^oC$ (Use $\log \ 2 = 0.3$,$\log \ 3 = 0.5$,$\log \ 5 = 0.7$,$K_b(NH_3) = 10^{-5}$)

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