Question

1) Calculate the change in pH when 3.00ml of .1M HCl(aq) is added to 100 ml of a buffer solution that is .1 M in NH3(aq) and .1M in NH4Cl(aq)

2)Calculate the change in pH when 3.0ml of .1 M NaOH (aq) is added to the original buffer solution

Answer 1

initially

0.1 M NH3 and 0.1 M NH4Cl

we know that

for basic buffers

pOH = pKb + log [ salt / base ]

also

pKb for ammonia is 4.75

so

pOH = pKb + log [NH4Cl/ NH3]

pOH = 4.75 + log [0.1 / 0.1]

pOH = 4.75

now

pH = 14 - pOH

so

pH = 14 - 4.75

pH = 9.25

now

3 ml of 0.1 M HCl is added

now

moles = conc x volume (L)

so

moles of HCl added = 0.1 x 3 x 10-3 = 0.3 x 10-3

moles of NH3 = 0.1 x 100 x 10-3 = 10 x 10-3

moles of NH4Cl = 0.1 x 100 x 10-3 = 10 x 10-3

now

the reaction is

NH3 + HCl ---> NH4Cl

we can see that

moles of NH3 reacted = moles of HCl added = 0.3 x 10-3

moles of NH4Cl formed = moles of HCl added = 0.3 x 10-3

so

finally

moles of NH3 = 10 x 10-3 - 0.3 x 10-3 = 9.7 x 10-3

moles of NH4Cl = 10 x 10-3 + 0.3 x 10-3 = 10.3 x 10-3

so

pOH = 4.75 + log [ 10.3 x 10-3 / 9.7 x 10-3 ]

pOH = 4.776

now

pH = 14 - 4.776

pH = 9.224

so

change in pH = 9.224 - 9.25

change in pH = -0.026

so

the pH is decreased by 0.026

2)

moles of NaOH added = 0.1 x 3 x 10-3 = 0.3 x 10-3

moles of NH3 = 0.1 x 100 x 10-3 = 10 x 10-3

moles of NH4Cl = 0.1 x 100 x 10-3 = 10 x 10-3

now

the reaction is

NH4+ + OH- ---> NH3 + H20

we can see that

moles of NH4+ reacted = moles of OH- added = 0.3 x 10-3

moles of NH3 formed = moles of OH- added = 0.3 x 10-3

so

finally

moles of NH3 = 10 x 10-3 + 0.3 x 10-3 = 10.3 x 10-3

moles of NH4+ = 10 x 10-3 - 0.3 x 10-3 = 9.7 x 10-3

so

pOH = 4.75 + log [ 9.7 x 10-3 / 10.3 x 10-3 ]

pOH = 4.724

now

pH = 14 - 4.724

pH = 9.276

so

change in pH = 9.276 - 9.25

change in pH = 0.026

so

the pH is increased by 0.026