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Grade: 12th pass
        
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.  4.7033
b.  5.7033
​​​c.  6.7033
​d.  8.7033
Ans ;- (b)  5.7033
Please explain in detail..............
one month ago

Answers : (1)

Arun
22031 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 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

one month ago
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