How many grams of solid ammonium bromide should be added to 0.500 L of a 0.135 M ammonia solution to prepare a buffer with a pH of 10.190? ka = 1.8Ã10-5

grams ammonium bromide = g.

0.7590 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.

Given that the base dissociation constant = 1.8*10⁻⁵

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

Given concentration of base = [base] = 0.135 M

pH = 10.190

pOH = 14 - pH = 14 - 10.190 = 3.81

So,

3.81 = 4.75 + log[acid]/0.135

[Acid] = 0.0155 M

Given that Volume = 0.5 L

So, Moles = Molarity * Volume

Moles = 0.0155 * 0.5 = 0.00775 moles

Molar mass of ammonium bromide = 97.94 g/mol

Mass = Moles * Molar mass = (0.00775 * 97.94) g = 0.7590 g