Ascorbic acid (vitamin C) is a diprotic acid, `H _(2) C _(6) H _(6) O _(6). [H^(+)] , pH and [C _(6)H _(6)O _(6) ]^(2-)` in a `0.10 M` solution of ascorbic acid will be respectively(Given `K _(a_1) = 6.8 xx 10 ^(-5) and K _(a_2) = 2. 7 xx 10 ^(-12)` for ascorbic acid).

Options

A. `3.6 xx 10^(-5)M, 2.58, 2.7 xx 10^(-12)M`
B. `2.6 xx 10^(-3)M, 5.18, 1.5 xx 10^(-10)M`
C. `2.6 xx 10^(-3) M, 2.58, 2.7 xx 10^(-12)M`
D. `1.5 xx 10^(-4)M, 3.67, 1.7 xx 10^(-14)M`

Answer

C

Explanation

`[H^(+)]` is determined by the first protic equilibrium, `H_(2)C_(6)H_(6)O_(6) hArr H^(+) + HC_(6)H_(6)O_(6)^(-)`
The mass action expression is : `K_(a_(1)) = 6.8 xx 10^(-5) = (x^(2))/(0.10)`
So, `x = [H^(+)] = 2.6 xx 10^(-3) M` and `pH = -log (2.6 xx 10^(-3)) = 2.58`
The concentration of the anion, `[HC_(6)H_(6)O_(6)^(-)]` , is given almost entirely by the second ionization equilibrium
`HC_(6)H_(6)O_(6)^(-) hArr H^(+) + C_(6)H_(6)O_(6)^(2-)`, for which the mass action espression is
`K_(a_(2)) = ([H^(+)][C_(6)H_(6)O_(6)^(2-)])/([HC_(6)H_(6)O_(6)^(-)]) = 2.7 xx 10^(-12)`
Using te value of x from the first setp above gives `2.7 xx 10^(-12) = ((2.6 xx 10^(-3))[C_(6)H_(6)O_(6)^(2-)])/((2.6 xx 10^(-3)))`
So, `[C_(6)H_(6)O_(6)^(2-)] = 2.7 xx 10^(-12) M`.