Question

A solution is prepared by dissolving MgCl2 (1.3 g/L) and NH4Cl (20.0 mg/L) into deionized water. The pH of the solution is adjusted to 8.85 using NaOH. What is the concentration of ammonia, NH3(aq), in this solution? The pKa of NH4+ is 9.24.

Answer 1

Magnesium chloride is readily soluble in water. It gives Mg²⁺ + 2Cl^-

MgCl₂(s) + H₂O ---> Mg²⁺(aq) + 2Cl^-(aq)

Ammonium chloride is also soluble in water and NH₄⁺ + Cl^-

NH₄Cl (aq) ----> NH₄⁺(aq) + Cl^-(aq)

NH₄⁺ is a weak acid with a pKa of 9.24.

The dissociation is:

NH₄⁺ -<--> NH₃ + H⁺

First calculate molarity of NH₄⁺

Molar mass: 53.491 g/mol

NH₄Cl (20.0 mg/L)

= 0.020/53.491 g/mol

= 0.000374 mol/l

NH₄⁺ = 0.000374 M

Use the Henderson-Hasselbalch equation and calculate concentration of ammonia

pH = pKa + log [(base)/(acid)]

8.850 = 9.24 + log [(base)/(acid)]

log [(base)/(acid)] = 8.850 - 9.24 =- 0.39

[(base)/(acid)] = 10^-0.9

           = 0.407: 1

= 407: 1000

NH₄⁺ = 0.000374 M

So, NH₃ = 0.000374 x 407/1000 mol/l

= 0.00015221 M = 1.5 x 10^-4 M

Concentration of ammonia is 1.5 x 10^-4 M