Question

Experiment VII: Buffers Lab Report ( 49 pts) I. Determination of the K, of acetic acid 1. Measured pH of the solution PH 2.40 < Calculate the [H3O*) at equilibrium for this solution. (include units) Tog [HT] = 10-2.40 H 0% 0.004 m. 3. Calculate the [CH,C00] at equilibrium for this solution. (include units) [CH.CO01 0.004 m 4. What is the [CH3COOH) at equilibrium for this solution? (include units) 700. [CH.COOH): 0.00lem 5. Based on these values, what the acid dissociation constant (Ka) of acetic acid? (include units) ka = 0.0040.004 0.006 x 0.0021 mol/l

Answer 1

The solution which contain both a weak acid (here acetic acid) and its conjugate base (here CH3COO-) , then it is called a Buffer. And pH of a buffer is calculated by Henderson Hasselbalch equation

pH = pKa + log [salt]/[acid]

Now given pH of the solution = 2.40

This the [H3O+] = 10^-pH = 10^-2.40 = 0.00398 M

The dissociation of CH3COOH is water is

CH3COOH + H2O ------------> CH3COO- + H3O+

Now the given [H3O+] is actually at equilibrium. Thus if we consider the initial concentration of  CH3COOH taken = xM

Then the ICE table will be

Reaction	CH3COOH	CH3COO-	H3O+
Initial	xM	0	0
Change	-0.00398 M	+0.00398 M	+0.00398 M
Equilibrium	xM-0.00398 M	0.00398 M	0.00398 M

Now Ka for the reaction will be-

Ka = [CH3COO-][H3O+] / [CH3COOH]

Now if the given [CH3COOH] at equilibrium = xM-0.00398 M = 0.006 M, then putting the values-

Ka = [CH3COO-][H3O+] / [CH3COOH]

Ka = [0.00398][0.00398] / [0.006]

= 0.00264

Now the actual Ka value for Acetic acid = 1.8x10^-5 = 0.000018

Thus the % error = (theoritical value - calculated value) / theoritical value * 100

= (0.000018 - 0.00264) / 0.000018 * 100

= [-0.002622 ] / 0.000018 * 100

=  [0.002622 ] / 0.000018 * 100

= 14566 %

This indicates that our calculated Ka value in completely wrong. Since the error should be within +/- 5%

Possible source of error- wrong calculation of initial concentration

- wrong calculation of pH

- wrong calculation of CH3COOH left at equilibrium