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Question

Calculate the amount of NH3 and molL1 required to premare a buffer solution of pH 9.0 when total concentration of buffering reagents is 0.6 mol L1.pKb for NH3=4.7,log2=0.30

A
0.4, 0.4
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B
0.2, 0.4
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C
0.4, 0.2
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D
0.2, 0.2
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Solution

The correct option is C 0.4, 0.2
pOH=logKb+log[Salt][base] 5=4.7+logabab=2 a=2b
Given a+b=0.6
2b+b=0.6
3b=0.6
Or b=0.2 mole, or 0.2×17=3.4g/L
a=0.4 mole
Or 0.4×53.5=21.4g/L
Thus [Salt] = 0.4M and [Base] = 4.8

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