Chemistry, 08.06.2021 03:20 avisconti571

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?

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Answer from: Jenniferlam12

Solution :

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

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

$K_a=\frac{[A^-].[H^+]}{[HA]}$

Let s calculate the concentration of HA and NaOH

$[HA] = \frac{^nH_A}{V}$

$=\frac{0.00285 \ mol}{0.04 \ L}$

= 0.07125 M

$[NaOH]= \frac{0.015L \times 0.0950 M}{V}$

$=\frac{0.001425 mol}{0.04L}$

= 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^+]

$[H^+] = 10^{-pH}$

$=10^{-6.5}$

$=3.16 \times 10^{-7}$

Calculating the value of K_a ,

$K_a=\frac{[A^-].[H^+]}{[HA]}$

$=\frac{0.0356 \times 3.16 \times 10^{-7}}{0.03565}$

$=3.16\times 10^{-7}$

Therefore the the value of K_a for the unknown acid is $3.16\times 10^{-7}$ .

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A sample of 0.2140 g of an unkown substance monoprotic acid was dissolved in 25.0 mL of water and ti...

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