Question

Calculate the change in pH when 3.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3 (aq) and 0.100 M in NH4CI(aq). Consult the table of ionization constants as needed ДрН Calculate the change in pH when 3.00 mL of 0.100 M N2OH is added to the original buffer solution. АрН -

Answer 1

Sol :-. By using Henderson-Hasselbalch equation :

pOH = pK_b + log [Salt]/[Acid]

pOH = pK_b of NH₃ + log [NH₄Cl]/[NH₃]

pOH = 4.75 + log 0.100 M/0.100 M

pOH = 4.75 + log 1

As log 1 = 0

So,

pOH = 4.75

and

pH = 14 - pOH

pH = 14 - 4.75

pH = 9.25

(a). pH after the addition of HCl

Number of moles of NH₄Cl = Molarity x Volume

= 0.100 M x 0.1 L

= 0.01 mol

Number of NH₃ = Molarity x Volume

= 0.100 M x 0.1 L

= 0.01 mol

and

Number of moles of HCl = Molarity x Volume

= 0.100 M x 0.003 L

= 0.0003 mol

ICF table is :

.........................NH₃................+................HCl -------------------> NH₄Cl

Initial ................0.01 mol..........................0.0003 mol..................0.01 mol

Change............-0.0003 mol......................-0.0003 mol................+0.0003 mol

Final...................0.0097 mol.......................0.0 mol........................0.0103 mol

again

By using Henderson-Hasselbalch equation :

pOH = pK_b + log [Salt]/[Acid]

pOH = pK_b of NH₃ + log [NH₄Cl]/[NH₃]

pOH = 4.75 + log 0.0103 /0.0097

pOH = 4.75 + 0.0261

So,

pOH = 4.78

and

pH = 14 - pOH

pH = 14 - 4.78

pH = 9.22

So, Change in pH = ΔpH = 9.22 - 9.25 = - 0.03 (Decreases)

-----------------------------

(b). After the addition of NaOH :-

Number of moles of NaOH = Molarity x Volume

= 0.100 M x 0.003 L

= 0.0003 mol

ICF table is :

.........................NH₄Cl..............+................NaOH -------------------> NH₃ ................+...........H₂O.......+.........NaCl

Initial ................0.01 mol..........................0.0003 mol..................0.01 mol

Change............-0.0003 mol......................-0.0003 mol................+0.0003 mol

Final...................0.0097 mol.......................0.0 mol........................0.0103 mol

again

By using Henderson-Hasselbalch equation :

pOH = pK_b + log [Salt]/[Acid]

pOH = pK_b of NH₃ + log [NH₄Cl]/[NH₃]

pOH = 4.75 + log 0.0097 /0.0103

pOH = 4.75 - 0.0261

So,

pOH = 4.72

and

pH = 14 - pOH

pH = 14 - 4.782

pH = 9.28

So, Change in pH = ΔpH = 9.28 - 9.25 = 0.03 (Increases)

-----------------------------