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$pK_{a1}$ and $pK_{a2}$ of $H_2CO_3$ are 6.38 and 10.26 respectively. The $pH$ of 1 M and 0.1 M $NaHCO_3$ are

(a) 8.32, 7.32

(b) 7.32, 8.32

(c) 8.32, 8.32

(d) 7.32, 7.32

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  • pH in this case is independent of concentration.
Answer: 8.32, 8.32
 
$pH = \frac{1}{2} [pK_{a1} + pK_{a2}] = \frac{1}{2}$ [6.38 + 10.26] = 8.32
Since, pH in this case is independent of concentration.
The pH of 1 M $NaHCO_3$ = pH of 0.1 M $NaHCO_3$ = 8.32

 

answered Nov 29, 2013 by mosymeow_1
 

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