Question

How many grams of solid ammonium chloride should be added to 1.50 L of a 0.145 M ammonia solution to prepare a buffer with a pH of 10.190 ?

Answer 1

Answer :

Problem related to basic buffer solution

Use of Henderson Hesselbatch equation

NH_{​​​​​3} + HCl ---> NH_{​​​​​4}​​​​Cl

pOH = pK_{​​​​​b} + log ([NH_{​​​​​4} Cl ]/[ NH_{​​​​​3}])

Given , pH = 10.190

So, pOH = 14 - pH = 14 - 10.19 = 3.81

Also, K_{​​​​​​b} of NH_{​​​​​3} (weak base) = 1.8 * 10^-5 (from Google)

pK_{​​​​​b} = - log (K_{​​​​​​b} ) = -log (1.8 * 10^-5 ) = 4.74

Given , Molarity of ammonia Solution (M) = 0.145 M

Volume = 1.5 L

Moles of NH_{​​​​​3} solution = Molarity * volume (L) = 0.145 * 1.5 = 0.2175 moles

[NH_{​​​​​3} ] = 0.2175 moles , [NH_{​​​​​4} Cl] = ?

Using Henderson Hesselbatch equation

pOH = pK_{​​​​​b} + log ([NH_{​​​​​4} Cl] / [NH_{​​​​​3} ] )

=> 3.81 = 4.74 + log ( [NH_{​​​​​4} Cl] / 0.2175 )

log ([ NH_{​​​​​4} Cl] / 0.2175 ) = 3.81 - 4.74 = -0.93

Taking antilog on both sides

[NH_{​​​​​4} Cl] / 0.2175 = 10^-0.93 = 0.1175

[NH_{​​​​​4} Cl] = 0.1175 * 0.2175 = 0.025 moles

Molar mass of NH_{​​​​​4} Cl = 14 + (4*1) + 35.5 = 53.5 g/mol

Mass of NH_{​​​​​4} Cl required = molar mass * moles

= 53.5 g/ mol * 0.025 moles = 1.3375 g

Mass of ammonium chloride required is 1.34 g