Question

Calculate the pH of a mixture that contains 0.20 M of HNO3 and 0.27 M of HC6H5O.

Answer 1

we know that

botH HN03 and HC6H5O are acids

so

both contribute [H+]

first consider HN03

we know that

HN03 is a strong acid

so 100% dissociation

HN03 ---> H+ + N03-

so

[H+] = [HN03] = 0.2

now

consider HC6H5O

it is a weak acid

we know that

for weak acids

[H+] = sqrt ( Ka x C)

given

Ka = 1.122 x 10-10

concentration (C) = 0.27

so

[H+] = sqrt ( 1.122 x 10-10 x 0.27)

[H+] = 5.5 x 10-6

now

total [H+] = 0.2 + (5.5 x 10-6)

total [H+] = 0.200005504

total [H+] = 0.2

now

we know that

pH = -log [H+]

so

pH = -log 0.2

pH = 0.6989

so

the pH of the solution is 0.7