Question

Ill give a medal!!
A buffer is created by dissolving 0.450 mol of sodium ascorbate and 0.500 mol ascorbic acid in enough water to create 1.00L of solution. The Ka of ascorbic acid is 7.9 x 10^-5.
A)what is the ph of the original buffer?

Answer 1

Well, we know that a buffer contains a weak acid and it's corresponding salt right?

The Ka gives us the acid dissociation constant which is a measure of the tendency that the acid has to give up it's proton. I believe ascorbic acid judging by the ka value because it's less than one means the following

\[k_{a} = \frac{ [H][A^{-}] }{ [HA]} = \frac{ [C_{6}H_{7}O_{6}][H^+] }{ [C_{6}H_{7}O_{6}] }\]

\[k_{A} < 1, means that~there~is~predominately~more~un-protinated~acid~at~equilibrium\]

so ascorbic acid, is a weak acid and is going to only give up a small % of H

\[C_{6}H_{8}O_{6} \rightarrow C_{6}H_{7}O_{6}^- + H^{+}\]

Answer 2

Well, this is a very long problem lol but let's get through it.

what we would need to do is I think we would have to make an ICE table.

like in our reaction

the concentration of the products will increase by x, and the reactants will decrease by x.

\[\frac{ products }{ reactants } = ka\]

\[\frac{ [+x][+x] }{ [M-x] } = K_{a}\]

we would need to find the concentration of the ascorbate ion


So we know that we're going to lose some ascorbaic acid right?

we know we have 0.45mol of ascorbic ion
and 0.5 mol of ascorbic acid to start.

we're given 1 liter

so that becomes

0.45 mol/L ascorbate
0.50 mol/L ascorbic acid


\[\frac{ [x][0.45+x] }{ [0.50M-x] } = 7.9*10^{-5}\]

you know that because this is a weak acid we know that there would be only a small amount of H+ so we can ignore the x on the bottom. for the reasons I said before.

\[\frac{ x(0.45+x) }{ 0.50 } = 7.9*10^{-5}\]


\[x^{2}+0.45x = 7.9*10^{-5}*0.50\]

you can actually solve this to get the concentration of [H+]

Answer 3

\[x^{2}+0.45x-3.95*10^{-5} = 0 \]

If you put this into wolfram-alpha you get an answer like this:

\[0.00087, -\log(0.00088) = 7.01\]