Question

A 20mL sol'n was composed of .20M NH3 and .25 M NH4Cl

a) determine the sol'n pH (ka= 5.6 x 10^-10

b) determine the pH after 1.00mL of 1.00M HCl was added

Answer 1

A 20mL sol'n was composed of .20M NH3 and .25 M NH4Cl

a) determine the sol'n pH (

ka= 5.6 x 10^-10,

pKa= -log Ka = - log 5.6 x 10^-10= 9.25

pH = pKa + log [NH3] / [NH4+]

pH = 9.25+ log 0.20 / 0.25

pH = 9.25+ log 0.80

pH = 9.25 – 0.097

pH = 9.183

b) determine the pH after 1.00mL of 1.00M HCl was added

Equation:
NH3 + HCl → NH4Cl
1mol NH3 reacts with 1 mol HCl to produce 1 mol NH4Cl .

Mole of NH3 = 20mL *0 .20M/1000 ml= 0.004 NH3
Mole of NH4+ = 20mL *0 .25M/1000 ml= 0.005 NH4+

Mole of HCl = 1.0ml /1000 *1.00M= 0.001 mol HCl

New mole of NH4+ = 0.005+0.001 = 0.006 mol NH4+

New mole of NH3 = 0.004- 0.001 = 0.003 mol NH3

Molarity = moles / volume in L

Total volume = 20+1.0 ml= 0.021L

Molarity of NH3 and NH4+:

NH4+ = 0.006/0.021=0.286M

NH3 = 0.003/0.021=0.143M

To calculate pH use the Henderson - Hasselbalch equation for a basic buffer:
pH = pKa + log [NH3] / [NH4+]

pH = 9.25+ log 0.143 / 0.286

pH = 9.25+ log 0.5

pH = 9.25 – 0.30

pH = 8.95