Acid-dissociation constants of two acids are listed in the table above. A 20. mL sample of a 0.10 M solution of each acid is titrated to the equivalence point with 20. mL of 0.10 M NaOH. Which of the following is a true statement about the pH of the solutions at the equivalence point?

(A) Solution I has a higher pH at the equivalence point because CH_3CO_2H is the stronger acid.
(B) Solution 1 has a higher pH at the equivalence point because CH_3CO_2H has the stronger conjugate base.
(C) Solution 1 has a lower pH at the equivalence point because CH_3CO_2H is the stronger acid.

The solution 1 has a higher pH at the equivalence point because CH3COOH it has a stronger conjugate base.

Explanation:

The strength of an acid is determined by the ease with which it releases hydrogen as ion in solution.

Given;

ka1 = 1.7× 10^-5

Ka2 = 1.0 ×10^0 = 1

So we need to calculate the concentration of [H+] released by each of these acids at equilibrium.

mole(CH3COO) = 0.1 × (20/1000)

= 0.002 mole

Mole(CF3COOH) = 0.002

After the titration; volume of solution = 40mL

[CH3COOH] = 0.002/0.04= 0.05 M

CH3COOH<--->CH3COO- + H3O+

0.05-x x. x

Ka = [H3O+][CH3COO-]/[CH3COOH]

Ka1 = x²/0.05-x

1.7 × 10^-5(0.05-x) = x²

8.5×10^-7 - 1.7 × 10^-5x = x²

x² + 1.7 × 10^-5x -8.5× 10^-7 = 0

We solve for using the quadratic formula

x = [1.7×10^-5 ±√(1.7×10^-5)²+4.1.(8.5×10^-7]/2

x = [ -0.000017±√3.4×10^-6]/2

= [-0.000017±0.001844]/2

x = [0.001844-0.000017]/2

x = 0.0009135

[H3O+] = 0.0009134 mol/dm³

pH = -log[H3O+] = 3.04

Similarly at equilibrium;

CF3COOH<---> CF3COO- + H3O+

0.05-x. x. x

Ka2=[CF3COO-][H3O+]/[CF3COOH]

1 = x²/0.05-x

0.05-x = x²

x² + x - 0.05= 0

x =[ - 1 ± √1 + (4)(1)(0.05)]/2

x = [-1 ± √1+0.2]/2

x =[ -1 ± 1.009950]/2

x = [1.009950-1]/2

x = 0.009950/2

x = 0.004975

[H3O+] = 0.004975

pH = 2.3

Seeing that that;

pHksolution1 > pHsolutoon2,

Hence solution 1 has a higher pH because it has a stronger conjugate base.