Calculate mass of ammonium chloride needed to prepare 1000 ml solution with pH value equal to 6.2.

General information:
Ammonium chloride (NH₄Cl) is a salt of ammonia (weak base). Ammonium chloride completely ionizes in solution to produce ammonim (NH₄⁺) and chloride (Cl^-) ions as shown in the following formula:

NH₄Cl_{(in aqueous solution)} à NH₄⁺_{(in aqueous solution)} + Cl^-_{(in aqueous solution)} 

Ammonium ions act as weak monoprotic acid in solution. It partially ionizes (dissociates) to produce ammonia (NH₃) and hydronium ions (H₃O⁺) as shown in the following formula:

NH₄⁺ + H₂O ↔ NH₃ + H₃O⁺

Ammonium chloride has a molar mass (molecular weight) equal to 53.49 g/ mole. Ammonia has Kb value equal to 1.74 × 10^-5  at 25 ºC.

Molar concentration of a weak monoprotic acid required to produce certain pH value in solution can be calculated from the following equations (see where it came from):

[H₃O⁺] = [H⁺]
C = ((2 × [H⁺] + Ka)² - Ka² ) ÷ (4 × Ka)

Where C is molar concentration of the monoprotic acid, [H⁺] is molar concentration of hydrogen (hydronium) ion, and Ka is the value of the acid ionization constant.

Calculations:
Step 1.  calculate Ka value of ammonium chloride from Kb value of ammonia as the following:
Ka × Kb = Kw
Ka = Kw/ Kb
Where Kw is the dissociation constant of water.
Ka = (1.00 × 10^-14)/ (1.74 × 10^-5)
Ka = 5.75 × 10^-10

Step 2. Calculate molar concentration required to produce solution pH equal to 3.2 using the above-mentioned equations. 
pH = - log [H⁺]

[H⁺] = 10^-pH

[H⁺] = 10^-pH = 10^-6.2 = 6.31 × 10^-7 mole/ liter

C = ((2 × 6.31 × 10^-7 + (5.75 × 10^-10))² - (5.75 × 10^-10)² ) ÷ (4 × (5.75 × 10^-10))

C = ((1.26 × 10^-6)² - (3.31 × 10^-19 )) ÷ (2.30 × 10^-9)

C = ((1.59 × 10^-12) - (3.31 × 10^-19)) ÷ (2.30 × 10^-9)

C= (1.59 × 10^-12) ÷ (2.30 × 10^-9)

C = 0.00069 mole/ liter

Use online pH calculator to check that 0.00069 mole/ liter ammonium chloride produces solution pH = 6.2. 

Step 3. Determine mass of ammonium chloride needed to prepare the solution.
Volume of solution in liters = 1000 ml × (0.001 liter/ ml) = 1.0 liter
Number of moles = concentration × volume = (0.00069 mole/ liter) × 1.0 liter = 0.00069 mole

Note: one mole ammonium chloride produces one mole ammonium ions in solution. Therefore, 0.00069 mole ammonium ions are produced from 0.00069 mole ammonium chloride.

Mass of ammonium chloride = number of moles × molar mass =0.00069 mole × (53.49 g/ mole) = 0.037 g

To make the preparation, dissolve 0.037 g ammonium chloride in freshly distilled water to make 1000 ml solution (using 1000 ml volumetric flask).

Related posts:
Prepare 500 ml acetic acid solution with a pH value equal to 2.7.
Calculate mass of hydrochloric acid (HCl) needed to prepare 100 ml solution with pH equal to 1.5
Calculate pH of 100 ml solution containing 0.05 g acetic acid and 0.06 g sodium acetate.
Calculate pH of 1 liter solution containing 0.5 g benzoic acid.