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InterviewSolution
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What is the `pH` of `7.0 xx 10^(-8)M` acetic acid. What is the concentration of un-ionsed acetic acid. `K_(a)` of `CH_(3)COOH = 1.8 xx 10^(-5)`. |
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Answer» a. In such dilute solutions, complete ionsiation of weak acid occurs, but ionisation of water must also be considered. Let `x [oversetΘ)OH] then [H_(3)O^(o+)] = x +(7.0 xx 10^(-8))` (From `H_(2)O` and acetic acid) `:.K_(w) - [H_(3)O^(o+)] [overset(Θ)OH] rArr (x +7.0 xx 10^(-8)) (x) = 10^(-14)`.or `rArr x^(2) +(7.0 xx 10^(-8))x - 10^(-14) = 0` `rArr x = 7.1 xx 10^(-8)M = [overset(Θ)OH]` Hence, `[H_(3)O^(o+)] (7.1 xx 10^(-8) +7.0 xx10^(-8))` `= 1.4 xx 10^(-7)M` `pH = - log (1.4 xx 10^(-7)) = 6.85` b. From charge balance (electroneutrality) Total negative charge = Total positive charge `[CH_(3)COO^(Θ)] +[overset(Θ)H] = [H_(3)O^(o+)]` `:. [CH_(3)COO^(Θ)] = [H_(3)O^(o+)] - [overset(Θ)OH]` `= (1.4 xx 10^(-7) - (7.1 xx 10^(-8))` c. `CH_(3)COOH +H_(2)O hArr CH_(3)COO^(Theta) +H_(3)O^(oplus)` `K_(a) = =([CH_(3)COO^(Θ)][H_(3)O^(o+)])/([CH_(3)COOH]) = 1.8 xx 10^(-15)` `[CH_(3)COOH] = ([CH_(3)COO^(Θ)][H_(3)O^(o+)])/(K_(a))` `= ((7 xx 10^(-8))(1.4 xx 10^(-7)))/(1.8xx10^(-5))` `= 5 xx 10^(-10)M` |
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