Answer :
Solution :
The equation is :
[tex]$HA (aq) + NaOH(aq) \rightleftharpoons NaA(aq) + H_2O(l)$[/tex]
The number of the moles of HA os 0.00285, and the volume is 25 mL.
15 mL of the 0.0950 M NaOH is added.
The total volume of a solution is V = 25 mL + 15 mL = 40 mL
The pH of the solution is 6.50
Calculating the [tex]K_a[/tex] of HA
[tex]$HA(aq) \rightleftharpoons A^-(aq)+H^+$[/tex]
[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]
Let s calculate the concentration of HA and NaOH
[tex]$[HA] = \frac{^nH_A}{V}$[/tex]
[tex]$=\frac{0.00285 \ mol}{0.04 \ L}$[/tex]
= 0.07125 M
[tex]$[NaOH]= \frac{0.015L \times 0.0950 M}{V}$[/tex]
[tex]$=\frac{0.001425 mol}{0.04L}$[/tex]
= 0.0356 M
[tex]$HA(aq) \ \ + \ \ NaOH(aq) \ \ \rightleftharpoons NaA(aq) \\ + \ \ H_2O(aq)$[/tex]
Initial conc. (M) 0.07125 M 0.0356 M 0 M
Change in conc. (M) -0.0356 M -0.0356 M + 0.0356 M
Equilibrium conc. (M) 0.03565 M 0 M 0.0356 M
Therefore, the concentration of HA and the NaA at the equilibrium are [HA] = 0.03565 M and [NaA]= 0.0356 M
0.0356 M of NaA dissociates completely into 0.0356 M [tex]Na^+[/tex] and 0.0356 M [tex]A^-[/tex]
Now for [tex][H^+][/tex]
[tex]$[H^+] = 10^{-pH}$[/tex]
[tex]$=10^{-6.5}$[/tex]
[tex]$=3.16 \times 10^{-7}$[/tex]
Calculating the value of [tex]K_a[/tex],
[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]
[tex]$=\frac{0.0356 \times 3.16 \times 10^{-7}}{0.03565}$[/tex]
[tex]$=3.16\times 10^{-7}$[/tex]
Therefore the the value of [tex]K_a[/tex] for the unknown acid is [tex]$3.16\times 10^{-7}$[/tex].