Question

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

Part A

0.130 M in HBr and 0.115 M in HCHO2

Part B

0.170 M in HNO2 and 8.0×10⁻² M in HNO3

Part C

0.190 M in HCHO2 and 0.23 M in HC2H3O2

Part D

4.0×10⁻² M in acetic acid and 4.0×10⁻² M in hydrocyanic acid

Answer 1

a)

[H+]total:

[H+] Br = 0.13 M

the ionization of HCHO2

HCOOH <-> H+ + HCOO-

Ka = [H+][HCOO-]/[HCOOH]

in equilbirum

[H+] = 0.13 + x

[HCOO- ] = x

[HCOOH] = 0.115- x

1.8*10^-4 = (x)(0.13+x) / (0.115- x)

0.00018*0.115 - 0.00018x = 0.13x + x^2

x^2 + (0.13018)x - 0.0000207 = 0

x = 1.588*10^-4

[H+] = 0.13 + 1.588*10^-4

[H+] =0.1301588 M

pH = -log(0.1301588 ) = 0.885

b)

similarly

[H+] from HNO3 = 0.08

HNO2 <- H+ + NO2-

K = 4*10^-4

4*10^-4 = (x)(0.08+x)/(0.17-x)

0.0004*0.17 - 0.0004x = 0.08x + x^2

x^2+ (0.08+0.0004)x - 0.000068 = 0

x = 8.37*10^-4

[H+} = 0.17 + 8.37*10^-4 = 0.170837

pH = -log(0.170837) = 0.7674