How many grams of solid ammonium chloride should be added to 2.00 L of a 0.273 M ammonia solution to prepare a buffer with a pH of 10.150? grams ammonium chloride = g. ka = 1.8Ã10-5

6.7970 g

Explanation:

Considering the Henderson - Hasselbalch equation for the calculation of the pOH of the basic buffer solution as:

pOH = pK + log[acid] / [base]

Where K is the dissociation constant of the base.

Base dissociation constant of the ammonia = 1.8*10⁻⁵

pK = - log (Kb) = - log (1.8*10⁻⁵) = 4.75

Given concentration of base = [base] = 0.273 M

pH = 10.150

pOH = 14 - pH = 14 - 10.150 = 3.85

So,

3.85 = 4.75 + log[acid]/0.273

[Acid] = 0.0347 M

Given that Volume = 2 L

So, Moles = Molarity * Volume

Moles = 0.0347 * 2 = 0.0694 moles

Molar mass of ammonium bromide = 97.94 g/mol

Mass = Moles * Molar mass = (0.00775 * 97.94) g = 6.7970 g