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`        Two buffers, (X) and (Y) of pH 4.0 and 6.0 respectively are prepared from acid HA and salt NaA, both the buffers are 0.50 M in HA. What would be the pH of the solution obtained by mixing equal volumes of the two buffers ?[ KHA=1.0×10-5].​a.  4.7033b.  5.7033​​​c.  6.7033​d.  8.7033Ans ;- (b)  5.7033Please explain in detail..............`
6 months ago

Arun
23389 Points
```							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 MNow 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 MNew 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
```
6 months ago
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