0.1 M, 60 mL NH_4Cl (K_h = 10^{-9}) is titr | vedclass

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(B) $1$. Initial moles of $NH_4Cl = 0.1 \ M 	imes 0.060 \ L = 0.006 \ mol$. $2$. Equivalence point occurs when moles of $NaOH$ added equal moles of $

Vedclass · Accepted Answer

$8.699$

Vedclass · Answer

(B) $1$. Initial moles of $NH_4Cl = 0.1 \ M 	imes 0.060 \ L = 0.006 \ mol$. $2$. Equivalence point occurs when moles of $NaOH$ added equal moles of $NH_4Cl$,i.e.,$0.006 \ mol$. $3$. Volume of $NaOH$ required for equivalence $= \frac{0.006 \ mol}{0.2 \ M} = 0.030 \ L = 30 \ mL$. $4$. At $\frac{1}{3}$ equivalence point,moles of $NaOH$ added $= \frac{1}{3} 	imes 0.006 = 0.002 \ mol$. $5$. Reaction: $NH_4^+ + OH^- ightarrow NH_3 + H_2O$. $6$. After reaction: $NH_4^+ = 0.006 - 0.002 = 0.004 \ mol$,$NH_3 = 0.002 \ mol$. $7$. Total volume $= 60 \ mL + 10 \ mL = 70 \ mL = 0.070 \ L$. $8$. This forms a basic buffer: $pOH = pK_b + \log \frac{[NH_4^+]}{[NH_3]} = -\log(K_b)$. $9$. Given $K_h = \frac{K_w}{K_b} = 10^{-9}$,so $K_b = \frac{10^{-14}}{10^{-9}} = 10^{-5}$. $10$. $pOH = 5 + \log \frac{0.004}{0.002} = 5 + \log(2) = 5 + 0.301 = 5.301$. $11$. $pH = 14 - pOH = 14 - 5.301 = 8.699$.

$0.1 \ M, 60 \ mL$ $NH_4Cl$ $(K_h = 10^{-9})$ is titrated against $0.2 \ M$ $NaOH$. What is the $pH$ at the $\frac{1}{3}$ equivalence point?

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