Question

A buffer is prepared by adding 15.0 g of ammonium chloride (NH4Cl) to 5.00L of 0.200 M ammonia (kb=1.8x10^-5). 1. What is the pH of the buffer solution? (assume the addition of ammonia does not change the volume) 2. What is the pH of the buffer after 90.0 mL of 2.00M HCl are added to the original buffer?

Answer 1

Mass of NH4Cl = 15.0 g

Molar mass of NH4Cl = 14.0067+ (4 1.0079 ) + 35.45 = 53.49 g / mol

No. of moles of NH4Cl = 15.0 g / ( 53.49 g / mol ) = 0.2804 mol

No. of moles of NH3 = [ NH3 ] Volume of solution in L

No. of moles of NH3 = 0.200 mol / L 5.00 L = 1.00 mol

We have Henderson's equation for basic buffer , pOH = pKb + log [ salt ]/ [ Base]

pOH = - log K b + log [ NH4Cl ] / [ NH3 ]

pOH = - log 1.8 10 -05 + log ( 0.2804 / 1.00 )

pOH = 4.74 + log ( 0.2804 / 1.00 )

pOH = 4.74 - 0.552

pOH = 4.188

We have relation, pH + pOH = 14

pH = 14 - pOH

pH = 14 - 4.188

pH = 9.812

ANSWER : pH of buffer solution = 9.81

2)

No. of moles of HCl = [ HCl​​​​​​​ ] Volume of solution in L

No. of moles of HCl =2.00 mol / L 0.090 L = 0.18 mol

Consider reaction of HCl with buffer solution.

NH3 + HCl NH4Cl

Let's use ICE table.

Concentration ( Moles )	NH3 + HCl NH4Cl
I	1.00	0.18	0.2804
C	-0.18	-0.18	+0.18
E	0.82	0.00	0.4604

Volume of buffer solution after addition of HCl is 5.00 L + 0.090 L = 5.09 L

We have Henderson's equation for basic buffer , pOH = pKb + log [ salt ]/ [ Base]

pOH = - log K b + log [ NH4Cl ] / [ NH3 ]

pOH = - log 1.8 10 -05 + log ( 0.4604 mol / 5.09 L  / 0.82 mol / 5.09 L )

pOH = 4.74 + log ( 0.4604 / 0.82 )

pOH = 4.74 - 0.251

pOH = 4.489

We have relation, pH + pOH = 14

pH = 14 - pOH

pH = 14 - 4.489

pH = 9.511

ANSWER : pH of buffer solution after addiiton of HCl = 9.51