InterviewSolution
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InterviewSolution
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When 900 ml of water is added in 100 ml of 0.01N HCl solution then pH of resultant solution will be: |
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Answer» Given that Volume of HCl = 100 ml = 0.1 L Normality of HCl = 0.1 N Now we can find Gram-equivalent of HCl = Normality × Volume = 0.1 × 0.1 = 0.01 So, Volume of solution = Volume of H2O + Volume of HCl = 900 + 100 = 1000 mL = 100 mL= 1 L Normality of HCl in resulting solution = 0.01/1 = 0.01 N pH = −log[H+] = −log(0.01) = 2 |
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