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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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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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