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?
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?
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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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