Question

Find the pH of each of the following solutions of mixtures of acids. No Ka is given.

Part A.) 0.125 M in HBr and 0.145 M in HCHO2 (Gave up, Answer pH= 0.903)

Part B.) 0.145 M in HNO2 and 9.0×10⁻² M in HNO3

Part C.) 0.190 M in HCHO2 and 0.23 M in HC2H3O2

Answer 1

Part A.) 0.125 M in HBr and 0.145 M in HCHO2 (Gave up, Answer pH= 0.903)

HBr strong acid and HCHO2 is weak acid and hence consider only HBr.

pH = - log [HBr] = -log [H+] = - log (0.125) = 0.903  

Part B.) 0.145 M in HNO2 and 9.0×10⁻² M in HNO3.

HNO2 is strong acid and HNO3 is strong acid.

but as the concentration of HNO3 is less and hence consider both ionization.

For HNO2, pKa = 3.347

pH = 0.5 * [pKa - log C]

or

pH = 0.5 * [3.347 - log (0.145)]

-log[H+] = 2.09

or

[H+] = 8.076 * 10^-3 M

thus

total concentration of H+ = ( 9.0×10⁻² + 8.076 * 10^-3) = 0.098 M

pH = -log (0.098) = 1.01

pH = 1.01

Part C.) 0.190 M in HCHO2 and 0.23 M in HC2H3O2:

for 0.190 M in HCHO2, pKa = 3.74

pH = 0.5 * [3.74 - log (0.190)] = 2.23

[H+] = 5.88 * 10^-3 M

and

0.23 M in HC2H3O2, pKa = 4.74

pH = 0.5 * [4.74 - log (0.23)] = 2.69

[H+] = 2.046 * 10^-3 M

thus

total concentration of H+ = (5.88 * 10^-3 + 2.046 * 10^-3 ) = 7.93 * 10^-3

pH = - log (7.93 * 10^-3) = 2.10

or

pH = 2.10