Match the following : Given pKa for CH3COOH=4.75,pKb, for NH4OH=4.75Column – IColumn – IIA) 60 ml of 0.1 M NaOH+40 ml of 0.05 M CH3COOHP) 7.00B) 50 ml of 0.1 M NH4OH+50 ml of 0.05M HClQ) 5.28C) 50 ml of 0.1 M NH4OH+50ml of 0.1M HCl R) 12.6D) 50 ml of 0.05 M CH3COOH+50ml of 0.05M NH4OH S) 9.25 T)Buffer solution

(A) NNaOH=4100=4×10−2 POH=1.4pH=14−1.4=12.6 ( excess strong base present)(B) POH=pKb+log[NH4Cl][NH4OH] =4.75+log2.5×10−32.5×10−3=4.75 (buffer)(C) [NH4Cl]=5×10−2M PH=12[logKb−logC−logKw] =12[−4.75−log 5×10−2+14]=5.28(Salt Hydrolysis)(D) [CH3COONH4]=2.5×10−2M pH=12[logKb−logKw−logKa] =12[−4.75+14+4.75]=7.00(Ammonium acetate is a buffer)

Match the following : Given pKa for CH3COOH=4.75,pKb, for NH4OH=4.75Column – IColumn – IIA) 60 ml of 0.1 M NaOH+40 ml of 0.05 M CH3COOHP) 7.00B) 50 ml of 0.1 M NH4OH+50 ml of 0.05M HClQ) 5.28C) 50 ml of 0.1 M NH4OH+50ml of 0.1M HCl R) 12.6D) 50 ml of 0.05 M CH3COOH+50ml of 0.05M NH4OH S) 9.25 T)Buffer solution

(A) NNaOH=4100=4×10−2 POH=1.4pH=14−1.4=12.6 ( excess strong base present)(B) POH=pKb+log[NH4Cl][NH4OH] =4.75+log2.5×10−32.5×10−3=4.75 (buffer)(C) [NH4Cl]=5×10−2M PH=12[logKb−logC−logKw] =12[−4.75−log 5×10−2+14]=5.28(Salt Hydrolysis)(D) [CH3COONH4]=2.5×10−2M pH=12[logKb−logKw−logKa] =12[−4.75+14+4.75]=7.00(Ammonium acetate is a buffer)

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Match the following :
Given $p K a$ for $C H_{3} C O O H = 4.75, p K b$ , for $N H_{4} O H = 4.75$
Column – I Column – II
A) $\begin{array}{l} 60 m l o f 0.1 M N a O H + 40 m l o f 0.05 M \\ C H_{3} C O O H \end{array}$ P) 7.00
B) $\begin{array}{l} 50 m l o f 0.1 M N H_{4} O H + 50 m l o f 0.05 M \\ H C l \end{array}$ Q) 5.28
C) $\begin{array}{l} 50 m l o f 0.1 M N H_{4} O H + 50 m l o f 0.1 M \\ H C l \end{array}$ R) 12.6
D) $\begin{array}{l} 50 m l o f 0.05 M C H_{3} C O O H + 50 m l o f 0.05 M \\ N H_{4} O H \end{array}$ S) 9.25
T) Buffer solution

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A – P, B – Q, C – R, D – S

A – R, B – ST, C – Q, D – PT

A – PT, B – QT, C –QT, D – PT

A – PT, B – R, C – Q, D – R

answer is D.

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Detailed Solution

(A) $N_{N a O H} = \frac{4}{100} = 4 \times 10^{- 2}$
$P O H = 1.4$
$p H = 14 - 1.4 = 12.6$ ( excess strong base present)
(B) $P O H = p K b + \log \frac{[N H_{4} C l]}{[N H_{4} O H]}$
$= 4.75 + \log \frac{2.5 \times 10^{- 3}}{2.5 \times 10^{- 3}} = 4.75 (b u f f e r)$
(C) $[N H_{4} C l] = 5 \times 10^{- 2} M$
$P H = \frac{1}{2} [\log K_{b} - \log C - \log K_{w}]$
$= \frac{1}{2} [- 4.75 - \log 5 \times 10^{- 2} + 14] = 5.28 (S a l t H y d r o l y s i s)$
(D) $[C H_{3} C O O N H_{4}] = 2.5 \times 10^{- 2} M$
$p H = \frac{1}{2} [\log K_{b} - \log K_{w} - \log K_{a}] = \frac{1}{2} [- 4.75 + 14 + 4.75] = 7.00 (Ammonium acetate is a buffer)$

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