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

the answer is c. nacl