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Question

A certain buffer solution contains X is twice the concentration of HX. If Ka for HX is 106, then pH of buffer is

A
7
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B
4.2
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C
6.3
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D
5.8
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Solution

The correct option is C 6.3
Given, Ka for HX is 106
Let the concentration of HX is x M
Then according to question,
concentration of its conjugate base [X]=2x M
Using the Henderson-Hasselbalch equation for weak acid:
pH=pKa+log[[conjugate base][acid]]
pH=log Ka+log[[X][HX]]
pH=log (106)+log[2xx]
pH=6+log 2
pH=6+0.301
pH=6.301

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