Question

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 1

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