InterviewSolution
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InterviewSolution
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A buffer solution contains `0.25M NH_(4)OH` and `0.3 NH_(4)C1`. a. Calculate the `pH` of the solution. b. How much `NaOH` should be added to `1L` of the solution to change `pH` by `0.6.K_(b) =2xx10^(-5)`. |
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Answer» Correct Answer - A::B i. Since it is basic buffer. Thus, `K_(b) = 2 xx 10^(-5)` `pK_(b) =- log (2 xx 10^(-5)) =- 0.3 +5 = 4.7` `pOH = pK_(a) + "log"(["Salt"])/(["Base"])` `= 4.7 + log ((0.3)/(0.25))` `= 4.7 + log (1.2)` `= 4.7 + log (12 xx 10^(-1))` `= 4.7 + log 2^(2) + log 3 - 1` `= 3.7 + 0.6 + 0.48 = 4.78` `pH = 14 - 4.78 = 9.22` ii. Rule `B B B:` In basic buffer (B),on addition of `S_(B)(B)`, the concentration of `W_(B)(B)` increase and that of salt decreases. On adding base, `pH` will increase and `pOH` will decrease. Let `xM NaOH` is added. `pH_("initial") = 9.22`, `pH_("new") = 9.22 + 0.6 = 9.86` `pOH_("new") = 14 - 9.86 = 4.14` `["Base"]_("new") = (0.25 +x)` `["Salt"]_("new") = (0.3 - x)` `:. pOH = pK_(b) + log ((0.3 -x)/(0.25 +x))` `4.14 = 4.74 + log ((0.3 -x)/(0.25 +x))` `4.14 - 4.74 = log ((0.3 - x)/(0.25+x))` `- 0.6 = log ((0.3 -x)/(0.25 +x)) "or log" ((0.25 +x)/(0.3-x)) = 0.6` `:. (0.25+x)/(0.3 -x) = "Antilog" (0.6) = 4` Solve for `x :` `x = 0.19` mol in `1L` of solution |
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