Question

8. Consider the titration of 30.0 mL of 0.100 M NH3 (K5= 1.8 x 10-) with 0.150 M HCl. Calculate the pH of the resulting solution after the 20.0 mL of HCl have been added. ₂NH2lag) + HClcag NHut cag) & crcag 30 mLX.1=3.

Answer 1

Given:

M(HCl) = 0.15 M

V(HCl) = 20 mL

M(NH3) = 0.1 M

V(NH3) = 30 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.15 M * 20 mL = 3 mmol

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

mol(NH3) = 0.1 M * 30 mL = 3 mmol

We have:

mol(HCl) = 3 mmol

mol(NH3) = 3 mmol

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

NH4+ here is strong acid

NH4+ formed = 3 mmol

Volume of Solution = 20 + 30 = 50 mL

Ka of NH4+ = Kw/Kb = 1.0E-14/1.8E-5 = 5.556*10^-10

concentration ofNH4+,c = 3 mmol/50 mL = 0.06 M

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

6*10^-2 0 0

6*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)*6*10^-2) = 5.774*10^-6

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

so, x = 5.774*10^-6 M

[H+] = x = 5.774*10^-6 M

use:

pH = -log [H+]

= -log (5.774*10^-6)

= 5.2386

Answer: 5.24