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Question

Calculate the ratio of CH3NH2 to CH3NH+3Cl required to create a buffer with pH = 10.14
Kb of CH3NH2=4.4×104
take log 2.27=0.356, log 0.313=0.504

A
0.5
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B
1.05
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C
0.31
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D
None of the above
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Solution

The correct option is C 0.31
Given, Kb of CH3NH2=4.4×104,pH=10.14
We need the Ka of the methylammonium ion:
We know at 25oC Ka×Kb=1014
Ka for CH3NH+3Cl=1014/4.4×104=2.27×1011
Using the Henderson-Hasselbalch equation:
pH=pKa+log[[conjugate base][acid]]
Substitute and solve for the base/acid ratio:
10.14=log2.27×1011+log[[conjugate base][acid]]
10.14=110.356+log[[conjugate base][acid]]
10.14=10.644+log[[conjugate base][acid]]
log[[conjugate base][acid]]=0.504
[conjugate base][acid]=antilog(0.504)
[conjugate base][acid]=0.313

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