Question

Q4, A 100.0mL sample of 0.10 M NH_{3 is titrated with} 0.10 M HNO₃ Determine the pH of the solution after addition of 100ml of HNO₃

The kb of NH₃ is 1.8x10^-5

Answer 1

Given:

M(HNO3) = 0.1 M

V(HNO3) = 100 mL

M(NH3) = 0.1 M

V(NH3) = 100 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.1 M * 100 mL = 10 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.1 M * 100 mL = 10 mmol

We have:

mol(HNO3) = 10 mmol

mol(NH3) = 10 mmol

10 mmol of both will react to form NH4+ and H2O

NH4+ here is acid

NH4+ formed = 10 mmol

Volume of Solution = 100 + 100 = 200 mL

use:

Ka = Kw/Kb

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/1.8*10^-5

Ka = 5.556*10^-10

concentration ofNH4+,c = 10 mmol/200 mL = 0.05 M

NH4+ + H2O -----> NH3 + H+

5*10^-2 0 0

5*10^-2-x x x

Ka = [H+][NH3]/[NH4+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.556*10^-10)*5*10^-2) = 5.27*10^-6

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

so, x = 5.27*10^-6 M

[H+] = x = 5.27*10^-6 M

use:

pH = -log [H+]

= -log (5.27*10^-6)

= 5.2782

Answer: 5.28