Question

Find the pH of each of the following solutions of mixtures of acids.

1)   7.0×10⁻² M in HNO3 and 0.180 M in HC7H5O2

2)   1.5×10⁻² M in HBr and 2.0×10⁻² M in HClO4

Answer 1

Q1.

HNO3 --> strong acid, will donate 100% so

[H+] =0.07 M

form the equilbirium, HC7H5O2 benzoic acid, has a pKa = 4.20

HA <-> H + A-

Ka = [H+][A-]/[HA]

initially

[H+] = 0.07

[A-] = 0

[HA] = 0.18

in equilibrium

[H+] = 0.07 + x

[A-] = 0 + x

[HA] = 0.18 - x

substitute

Ka = 10^-pKa = 10^-4.20

6.30*10^-5 = x*(0.07 + x) / ( 0.18 - x)

solve for x

(6.30*10^-5)*(0.18) = 0.07x + x^2

x^2 + 0.07x - 0.00001134 = 0

x = 1.616*10^-4

pH = -log( 1.616*10^-4) = 3.79

Q2.

HBr = strong acid, HClO4 = strong acid

so

[H+] total = [HBr] + [HClO4] = 1.5*10^-2 + 2*10'^-2 = 3.5*10^-2

[H+] = 0.035

pH = -log(0.035) = 1.4559