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The `pH` of a solution containing `0.1mol` of `CH_(3)COOH, 0.2 mol` of `CH_(3)COONa`,and `0.05 mol` of `NaOH` in `1L. (pK_(a) of CH_(3)COOH = 4.74)` is:A. `5.44`B. `5.20`C. `5.04`D. `4.74` |
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Answer» Correct Answer - A `{:(,CH_(3)COOH+,NaOHrarr,CH_(3)COONa+,H_(2)O),("Initial mol",0.1,0.05,0,0),("Final mol",(0.1-0.05),(0.05-0.05),0.05,-),(,=0.05,=0,,):}` Total moles of `Ch_(3)COONa = 0.2 + 0.05 = 0.25`. `:. [CH_(3)COONa] = (0.25mol)/(1L) = 0.25M` Moles of `CH_(3)COOH` left `= (0.1 -0.05) = 0.05` `:. [CH_(3)COOH] = (0.05mol)/(1L) = 0.05M` Thus, acidic buffer is formed `pH = pK_(a) + "log"(["Salt"])/(["Acid"])` `= 4.74 + log ((0.25)/(0.05))` `= 4.74 + log 5 = 4.74 + 0.7 = 5.44`. |
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