Question

U UI .500 M (HAC) = 0.5001). WEILS Luery Successive dilution, mean successive dilution, measure the corresponding to love centrations of acetic acid. Enter the measured pe values in Table 13. i below Table 13.1 - Ionization Constant of Acetic Acid (Acled (LAC) e ka I ionit pia zation "Recipe" for preparing Trial) initial concentration of HAC [Hic4 pH | [H] [HAC] = 0.500 0.500 2.30 Dilute 100 mL of (1) to 200 mL 257 3 Dilute 200 mL of (2) to 500 mL 2.36 Dilute 100 mL of (3) 90 to 200 mL Dilute 200 mL of (4) to 500 ml Before leaving the lab, calculate all of the required quantities for Trials 1 and 2. Percent ionization (last column) is given by the formula:

Answer 1

Reaction:

CH3COOH (aq) $\rightleftharpoons$ CH3COO-(aq) + H+(aq)

HAc = CH3COOH

Ac- = CH3COO-  

ICE table :

	[HAc]	[Ac-]	[H+]
I	a	0	0
C	-x	+x	+x
E	a-x	x	x

Equilibrium constant for this reaction, Ka  is given by

Ka = [Ac-][H+] / [HAc]

= x2 / (a-x)

pH = - log[H+]

pKa = - logKa

% ionisation = (amount of ionised acid / total amount) * 100

amount of ionised acid = [HAc-]

1. Given, [HAc] = 0.5 M = a

pH = 2.30

So, [H+] = 10^(-pH)

= 0.005 M = x

From ICE table we know that [H+] at equilibrium is equal to [Ac-] at equilibrium.

[Ac-] = 0.005 M

[HAc] at equilibrium = a-x

= 0.495 M

Ka = x2 / (a-x)

= 0.0052 / 0.495

= 5.05 * 10^(-5)

pKa = - logKa

= 4.30

% ionization = (x / a )*100

= 1.00 %

Similarly

Trail	[HAc]o	pH	[H+]	[Ac-]	[HAc]eq	Ka	pKa	%ionization
2	0.25	2.57	0.0027	0.0027	0.2473	2.95*10^(-5)	4.53	1.08 %
3	0.01	2.86	0.0014	0.0014	0.0086	2.28*10^(-4)	3.64	14.00 %
4	0.25	2.92	0.0012	0.0012	0.2488	5.75*10^(-6)	5.24	0.48 %
5	0.01	3.18	0.0007	0.0007	0.0093	5.27*10^(-5)	4.28	7.00 %

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