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pH of a solution of a strong acid is 5.0. What would be the pH of the solution obtained after diluting the given solution a 100 times ? |
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Answer» Solution :Given that, pH = 5 `[H^+]=10^(-5) "mol L"^(-1)` On diluting the solution 100 TIMES `[H^+]=10^(-5)/100= 10^(-7) "mol L"^(-1)` On calculating the pH using the equation `pH = -log [H^+]`, value of pH comes out to be 7. It is not POSSIBLE. This indicates that solution is very dilute. Hence, Total `H^+` ion concentration = `H^+` ions from ACID + `H^+` ion from water `[H^+]=10^(-7)+10^(-7)=2XX10^(-7)` M pH=-log `[2xx10^(-7)]` pH = 7 - 0.3010 = 6.699 |
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