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A buffer solution was made by adding `15.0 g` of `CH_(3) COOH` and `20.5gCH_(3)COONa`. The buffer is diluted to `1.0L`. a. Calculate the `pH` of solution. b. What will be the change in `pH` if `10.0mL` of `1.0 M HC1` is added to it. Given: `pK_(a) of CH_(3)COOH = 4.74, log ((13)/(12)) = 0.035` |
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Answer» Correct Answer - A::B::C::D i. Mw of `CH_(3)COOH = 60 g mol^(-1)` Mw of `CH_(3)COONa = 82g mol^(-1)` `["Salt"] = (20.5)/(82) = 0.25M (V = 1L)` `["Acid"] = (15.5)/(60) = 0.25M (V = 1L)` `pH = pK_(a) + log.(["Salt"])/(["Acid"])` `= 4.74 + log.([0.25])/([0.25]) = 4.74` ii. Rule `A A A:` In acidic buffer `(A)` on adding of `S_(A)(A)`, the concentration of `W_(A)(A)` increases and that of salt decrease. Moles of `HC1 = (10xx1M)/(1000) = 10^(-2)M = 0.01M` `[Acid] = 0.25 + 0.01 = 0.26M` `["Salt"] = 0.25 - 0.01 = 0.024` `pH_("new") = pK_(a) + "log" [("Salt")/("Acid")] = pK_(a) + log ((0.24)/(0.26))` `= pK_(a) + log ((12)/(13)) = pK_(a) - log ((13)/(12))` `= 4.74 - 0.035 = 4.705` Change in `pH:` `DeltapH = pH_("new") - pH_("inital")` `= 4.705 - 4.74 =- 0.035` |
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