Question

A sample of 0.2140 g of an unkown substance monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.950 M NaOH. The acid required 27.4 mL of base to reach the equivalence point. After 15.0 mL of base had been added in the titration, the pH was found to be 6.50. What is the Ka for the unknown acid?

Answer 1

The equation is :

`$HA (aq) + NaOH(aq) \rightleftharpoons NaA(aq) + H_2O(l)$`

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
`K_a` of HA

`$HA(aq) \rightleftharpoons A^-(aq)+H^+$`

`K_a=([A^-].[H^+])/([HA])`

Let s calculate the concentration of HA and NaOH

`$[HA] = (^nH_A)/(V)$`

`$=(0.00285 \ mol)/(0.04 \ L)$`

= 0.07125 M

`$[NaOH]= (0.015L * 0.0950 M)/(V)$`

`$=(0.001425 mol)/(0.04L)$`

= 0.0356 M

`$HA(aq) \ \ + \ \ NaOH(aq) \ \ \rightleftharpoons NaA(aq) \\ + \ \ H_2O(aq)$`

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
`Na^+` and 0.0356 M
`A^-`

Now for
`[H^+]`

`$[H^+] = 10^(-pH)$`

`$=10^(-6.5)$`

`$=3.16 * 10^(-7)$`

Calculating the value of
`K_a`,

`K_a=([A^-].[H^+])/([HA])`

`$=(0.0356 * 3.16 * 10^(-7))/(0.03565)$`

`$=3.16* 10^(-7)$`

Therefore the the value of
`K_a` for the unknown acid is
`$3.16* 10^(-7)$`.