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Standard XII

Chemistry

Henderson Hassleback Equation

To 1 L soluti...

Question

To 1L solution containing 0.1 mol each of $N H_{3}$ and $N H_{4} C l, 0.05$ mol of NaOH is added. The change in pH will be $(p K_{b} f o r N H_{3} = 4.74)$

-0.48

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0.48

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0.30

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-0.30

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Solution

The correct option is B
0.48

$N H_{4} C l + N a O H \to N a C l + N H_{3} + H_{2} O M o l e s o f N H_{4}^{+} l e f t = 0.1 - 0.05 = 0.05$
Total moles of $N H_{3} = 0.1 - 0.05 = 0.15$
$(p O H)_{1} = p H_{b} + l o g \frac{[S a l t]}{[B a s e]} = p K_{b} + l o g \frac{0.1}{0.1} = p K_{b}$
$(p O H)_{2} = p H_{b} + l o g \frac{0.05}{0.15} = p K_{b} - l o g 3$
Change in $p O H = (p O H)_{2} - (p O H)_{1} = - l o g 3$
Change in pH = log3 = 0.48

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To 1L solution containing 0.1 mol each of NH3 and NH4Cl,0.05 mol of NaOH is added. The change in pH will be (pKb for NH3=4.74)

The correct option is B 0.48 NH4Cl+NaOH→NaCl+NH3+H2OMolesofNH+4left=0.1−0.05=0.05 Total moles of NH3=0.1−0.05=0.15 (pOH)1=pHb+log[Salt][Base]=pKb+log0.10.1=pKb (pOH)2=pHb+log0.050.15=pKb−log 3 Change in pOH=(pOH)2−(pOH)1=−log 3 Change in pH = log3 = 0.48

To 1L solution containing 0.1 mol each of $N H_{3}$ and $N H_{4} C l, 0.05$ mol of NaOH is added. The change in pH will be $(p K_{b} f o r N H_{3} = 4.74)$