Answer: 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.
