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The pH of a solution containing `0.20M CH_(3)COOH` and `0.30M CH_(3)COONa` isA. `2.89`B. `4.92`C. `5.04`D. `3.89` |
Answer» Correct Answer - B Since we are dealing with a mixture of a weak acid and its salt with a strong base, we can calculate the pH of the solution by using the Henderson-Hasselbalch equation: `pH=pK_(a)+"log"(C_("salt"))/(C_("acid"))` In this case, we need to calculate `pK_(a)` and the acid first: `pK_(a)= -logK_(a)` `= -log(1.8xx10^(-5))` `=4.74` Now, we substitute the value of `pK_(a)` and concentrations of the acid and its salt: `pH=pK_(a)+"log"(C_("salt"))/(C_("acid"))` `=4.74+"log"(0.30)/(0.20)` `=4.74+log(1.5)` `=4.92` |
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