Question

How many grams of solid ammonium chloride should be added to 1.50 L of a 0.202 M ammonia solution to prepare a buffer with a pH of 9.880 ? grams ammonium chloride

Answer 1

Solution the first image :-

NH3(in water,NH4OH ) and NH4Cl(NH4+) is a buffer solution since there is difference of one OH-.Kb of ammonia is 1.8*10-5.

pOH = 14 - pH = 14 -  9.88 = 4.12, pKb = -log10[Kb] = -log10[1.8*10-5] = 4.74

For buffer solution - pOH = pKb + log10{ [Salt] / [Base] }

For above solution :-

pOH = pKb + log10{ [NH4+] / [NH4OH] } ⇒ 4.12 = 4.74 + log10{ [NH4+] / 0.202 } ⇒ [NH4+] = 0.048M

So concentration of NH4Cl required is 0.048M.

Moles of NH4Cl required = Molarity * Volume of solution(in Litres) = 0.048 * 1.5 = 0.0726 moles

Mass of  NH4Cl required = Moles of NH4Cl required * Molecular weight of NH4Cl =  0.0726* 53.49 = 3.887g