Question

A 21.0 mL sample of 0.291 M ammonia, NH3, is titrated with 0.210 M hydrobromic acid. The pH before the addition of any hydrobromic acid is

Answer 1

Only NH3 is present

Kb of NH3 = 1.8*10^-5

NH3 dissociates as:

NH3 +H2O -----> NH4+ + OH-

0.291 0 0

0.291-x x x

Kb = [NH4+][OH-]/[NH3]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.8*10^-5)*0.291) = 2.289*10^-3

since c is much greater than x, our assumption is correct

so, x = 2.289*10^-3 M

So, [OH-] = x = 2.289*10^-3 M

use:

pOH = -log [OH-]

= -log (2.289*10^-3)

= 2.6404

use:

PH = 14 - pOH

= 14 - 2.6404

= 11.3596

Answer: 11.36