Answer:
0.311 M
Explanation:
HA, our weak acid, reacts with NaOH by the following equation given below:
[tex]HA (aq) + NaOH (aq)\rightarrow NaA (aq) + H_2O (l)[/tex]
This may also be represented by the net ionic equation for this acid-base reaction:
[tex]HA (aq) + OH^- (aq)\rightarrow A^- (aq) + H_2O (l)[/tex]
During the titration up to the equivalence point, we would have some HA remaining and some [tex]A^-[/tex] produced. That said, we have a weak acid and its conjugate base forming a buffer. We may apply the Henderson-Hasselbach equation for buffers here:
[tex]pH = pK_a + log(\frac{[A^-]}{[HA})[/tex]
At midpoint, half of HA reacts to produce half of [tex]A^-[/tex], meaning [tex][HA] = [A^-][/tex] when half of equivalence volume of NaOH is added.
This implies that:
[tex]pH = pK_a[/tex] at midpoint. We may then find the acid ionization constant for this acid:
[tex]K_a = 10^{-pH} = 10^{-4.102} = 7.91\cdot 10^{-5}[/tex]
The initial molarity has to be found now. We know that the initial pH is:
[tex]pH = 2.308[/tex]
Since [tex]pH = -log[H_3O^+][/tex], then [tex][H_3O^+] = 10^{-pH}[/tex]
Using an ICE table, for some initial molarity of acid, [tex]c_o[/tex], we may write the equilibrium constant expression as:
[tex]K_a = \frac{[H_3O^+]^2}{c_o - [H_3O^+]}[/tex]
Solve for the initial molarity of HA:
[tex]c_o = \frac{[H_3O^+]^2}{K_a} + [H_3O^+] = \frac{10^{-2pH}}{K_a} + 10^{-pH} = \frac{10^{-2\cdot 2.308}}{7.91\cdot 10^{-5}} + 10^{-2.308} = 0.311 M[/tex]
