Question

Calculate the thoeretical pH of Solution B (10mL of 0.2M acetic acid and 10 mL of 0.1M sodium hydroxide). Then calculate the theoretical pH after an addition of 1 mL of 0.0992M of HCl. Initial pH: 4.66, pH after addition of acid: 4.66

Calculate the theoretical pH of solution B (10 mL of 0.2M sodium acetate and 10 mL of 0.1M HCl). Then calculate the theoretical pH after an additon of 1 mL of 0.997M of NaOH. Initial pH: 4.81, pH after: 486

Answer 1

1.

mmoles of acetic acid = 10× 0.2 = 2

mmoles of NaOH added = 10×0.1= 1

BCA table is

	Acetic acid	NaOH	Sodium acetate
Before	2	1	0
Change	-1	-1	+1
After	1	0	1

Theoretical pH can be calculated by

Henderson-Hasselbalch equation

pH = pKa + log(mmoles of acetate ion/acetic acid).

Or, pH = 4.75 + log(1/1)

Or, pH = 4.75.

Initial pH = 4.75

Now, when 1 mL 0.0992 M HCl is added.

mmoles of HCl = 1×0.0992 = 0.0992

HCl + CH₃COONa $\to$ CH₃COOH + NaCl

Now, BCA table is

mmoles	CH3COO-	HCl	CH3COOH
Before	1	0.0992	1
Change	-0.0992	-0.0992	+0.0992
After	0.9008	0	1.0992

New pH after addition of HCl

pH = 4.75+ log(0.9008/1.0982)

Or, pH = 4.75 - 0.09 = 4.66.

2.

mmoles of sodium acetate = 10×0.2= 2

mmoles of HCl = 10×0.1= 1.

HCl + CH₃COONa $\to$ CH₃COOH + NaCl.

BCA table is

	CH3COO-	HCl	CH3COOH
Before	2	1	0
Change	-1	-1	+1
After	1	0	1

pH = pKa + log (mmoles of acetate ion/mmoles of acetic acid).

Or, pH = pKa + log(1/1)

Or, pH = pKa

Now, when, 1 mL 0.997 M NaOH

mmoles of NaOH

= Volume × molarity

= 1× 0.997

= 0.997 .

BCA table is

	CH3COOH	NaOH	CH3COONa
Before	1	0.997	1
Change	-0.997	-0.997	+0.997
After	0.003	0	1.997

pH = 4.75 + log (1.997/0.003)

= 4.75 + 2.82 = 7.57.