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We have, pH for buffer as,
pH = – log Ka + log{[Salt]/[Acid]}
Case I : 4 = – log (1.0×10-5.) + log{[Salt]/(0.5)}
Therefore, log {[Salt]/(0.5)} = – 1
Or, [Salt] = 0.1×0.5 = 0.05 M
Case II : 6 = – log 1.0×10-5 + log{[Salt]/(0.5)}
Therefore, log {[Salt]/(0.5)} = 1
Therefore, [Salt] = 10×0.5 = 5 M
Now the two buffer[(I. NaA = 0.05 M & HA = 0.5 M) and (II. NaA = 5 M & HA = 0.5 M)] are mixed in equal proportion.
New concentration of NaA in mixed buffer = (0.05×V + 5×V)/2V = 5.05/2 M
New concentration of HA in mixed buffer = (0.5×V + 0.5×V)/2V
= 0.5 M
Therefore, pH = – log 1.0×10-5 + log{[5.05/2][0.5]}
= 5 + 0.7033
= 5.7033
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