Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0145 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid.

Answer 1

we know that

pH = -log [H+]

given

pH = 2.53

so

-log [H+] = 2.53

[H+] = 2.95121 x 10-3

now

let the monoprotic acid be HA

now

HA ---> H+ + A-

initially 0.0145 M HA is present

now

at equilirbium

[HA] = 0.0145 - y

[H+] = y

[A-] = y

we obtained

[H+] = 2.95121 x 10-3

so

y = 2.95121 x 10-3

so

[HA] = 0.0145 - ( 2.95121 x 10-3)

[HA] = 0.0115488

now

consider the equation

HA ---> H+ + A-

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

so

Ka = [ 2.95121 x 10-3 ] [2.95121 x 10-3 ] / [0.0115488]

Ka = 7.54 x 10-4

so

the value of Ka is 7.54 x 10-4