InterviewSolution
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InterviewSolution
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A solution of weak acid HA was titrated with base NaOH. The equivalent point was reached when 40 mL. Of 0.1 M NaOH has been added. Now 20 mL of 0.1 M HCl were added to titrated solution, the pH was found to be 5.0 What will be the pH of the solution obtained by mixing 20 mL of 0.2 M NaOH and 20 mL of 0.2 M HA?A. `7`B. `9`C. `10`D. `11` |
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Answer» Correct Answer - B `HA + NaOH rarr NaA + H_(2)O` milli moles of salt `NaA` or `A^(-) = 40 xx 0.1 = 4` `{:("Now",A^(-)+,H^(+)rarrHA),("Initial milli moles",4,2),("Final milli moles",2,-2):}` Acidic buffer solution is formed and `[A^(-)] = [HA]` `pH = pK_(a) + "log" ([A^(-)])/([HA]) rArr pK_(a) = 5` Now `HA + NaOH rarr NaA + H_(2)O`, hydrolysis of `A^(-)` will takes place `[NaA] = ("milli moles of acid")/("total volume") ` `= (20 xx 0.2)/(20 + 20) = 0.1` `pH = 1/2 (pK_(w) + pK_(a) + "log" C) = (1)/(2) [14+5-1] = 9` |
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