1.

A solution is found to contain 0.63 g of nitric acid per 100 ml of solution. What is the PH of solution if acid is completely dissociated? 

Answer»

M=n/V 

=w2/M2 x V 

= 0.63x1000/63 mol-1L-1 

=0.01 molL-1 

[H+ ] =[HNO3] =10-2 

 PH = - log(H+

 = - log(10-2

 = -(-2)log10 

 PH =2



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