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Question

Two buffers, (X) and (Y) of pH 4.0 and 6.0 respectively are prepared from acid HA and the salt NaA. Both the buffers are 0.50 M of HA. What would be the pH of the solution obtained by mixing equal volumes of the buffers ? KHA=1.0×105

A
8.4
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B
4.3
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C
5.7
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D
7.5
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Solution

The correct option is C 5.7
Henderson equation is given by pH=log10Ka+log10[Salt][Acid]
When pH=4
4=log101.0×105+log10[Salt]0.5
4=5+log10[Salt]0.5
log10[Salt]0.5=1
[Salt]=0.1×0.5=0.05
When pH=6,
6=log10(1.0×105)+log10[Salt]0.5
6=5+log10[Salt]0.5
log10[Salt]0.5=1
[Salt]=10×0.5=5 M
Suppose V litre each of both buffers are mixed. The concentration of the salt in this solution will be:
[Salt]=(0.05×V)+(5×V)2V=5.052 M
Concentration of [HA] in mixed buffer
=(0.5×V)+(0.5×V)2V=0.5 M
pH=log10(1.0×105)+log10[5.052]0.5
=5+0.7033=5.7033

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