Question

2.) Describe and show your calculations how you will make a 50 mL solution of an ammonia buffer where pH = 10.07 and [NH3] + [NH4+] = [NH4]T = 10.2 M using concentrated ammonia (28.0% by weight) and ammonium chloride.

Answer 1

Solution:

Given: pH = 10.07
or, pOH = 14 - 10.07 = 3.93
or, [OH-] = 10^3.93 = 8511.38 M

NH₃ + H₂O <------> NH₄⁺ + OH^-

Kb = [NH₄⁺][ OH^- ]/[ NH₃]
1.8 x 10^-5 = (x) 1 x 10^3.93 /(12.2-x)       (For NH₃, Kb =1.8 x 10^-5, [NH₃] =10.2-[NH₄⁺])

1.8 x 10^-5 = (x) 1 x 10^3.93 /12.2            (since, kb for NH₃ is very small, 12.2–x =12.2)
x = 12.2x1.8x10^-5/10^3.93

x = 21.96x10^-8.93 = 2.19x10^-7.93

or, [NH₄⁺] = 2.19x10^-7.93 M

Now, moles of [NH₄⁺] in 50 mL solution can be calculated as follows: 2.19x10^-7.93 M=2.19x10^-7.93 mol/L

2.19x10^-7.93 mol/L x 0.05 L = 0.1095x10^-7.93 mol
Molecular weight NH₄Cl = 53.5 g/mol
0.1095x10^-7.93 mol x 53.5 g/mol = 6.88x10^-8 g NH₄Cl

Also, [NH₃] =10.2-[NH₄⁺] = 10.2 – 2.19x10^-7.93 ~ 10.2 M

10.2M NH₃ solution = 10.2 moles/L

10.2 moles/L x 0.05 L = 0.51 mol NH₃

Molecular weight NH₃ = 17 g/mol

0.51 mol x17g/mol = 8.67 g NH₃

But, concentrated ammonia contains 28.0 % NH₃ by weight

Therefore, 28 g NH₃ = 100 mL NH₃

            8.67 g NH₃ = 100x8.67/28 mL NH₃ = 30.96 mL NH₃

Therefore, 30.96 mL NH₃ has to be diluted with distilled water to a total volume of 50 mL to have the buffer solution of the aforesaid pH.