Calculate the amount of NH3 and molL−1 required to premare a buffer solution of pH 9.0 when total concentration of buffering reagents is 0.6 mol L−1.pKbforNH3=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+logab⇒ab=2∴a=2b Given a+b=0.6 2b+b=0.6 ∴3b=0.6 Or b=0.2mole,or0.2×17=3.4g/L ∴a=0.4mole Or 0.4×53.5=21.4g/L Thus [Salt] = 0.4M and [Base] = 4.8