Question

ble 18.2 (continued) 3.2D Solutions Tested 0. 10 M HC HO, (HI(M) [C,H,O, (M) K I Percent Dissociation 0.010 M HC,H,O, 0.0010 M HC,H,O, pH 2.92 B39 579 Question Write the net lonic equation describing the dissociation of acetic acid in aqueous solution HC₂ H₂O₂ (aq) & At(aq) + CH32 Cag) Question Calculate (H) and (C,H,O," for each concentration above, showing a sample cal- culation here. Question Write the equilibrium constant expression for acetic acid, and calculate a K value and an apparent percent dissociation for each concentration, showing sample calculations here. estion What is the effect of dilution on the percent dissociation? Why should this behavior be expected?

I need help with how to get concentrations of H+ and C2H3O2, K, and percent dissociation of each solution tested. Also the K expression. Please explain!

Answer 1

Lets start the problem by answering how to calculate the H+ concentration. You are given the pH values, so we can use the formula pH = -log[H+] and rearrange it to calculate the [H+].         

                                                           [H+] = 10 -pH

i. pH = 2.92 : [H+] = 10 -pH = 10 -2.92   = 0.0012 M

ii. pH = 3.34: [H+] = 10 -pH = 10 -3.34 = 4.57x 10-4 M

iii. pH = 3.74: [H+] = 10 -pH = 10 -3.74 = 1.82 x 10-4 M

From the net ionic equation, we can see that [H+] and [C2H3O2-] are in 1:1 ratio which means concentration of the acetate ion is same as the concentration of hydrogen ion / hydronium ion.

i. pH = 2.92 : [H+] = 10 -pH = 10 -2.92   = 0.0012 M = [C2H3O2-]

ii. pH = 3.34: [H+] = 10 -pH = 10 -3.34 = 4.57x 10-4 M = [C2H3O2-]

iii. pH = 3.74: [H+] = 10 -pH = 10 -3.74 = 1.82 x 10-4 M = [C2H3O2-]

The equilibrium concentration expression for an equation is given by K = [products] / [reactants]

From the ionic equation we can write K = [H+] [C2H3O2-] / [CH3COOH]

i. K = [0.0012 M][0.0012 M] / [0.1 M] = 1.44 x 10-5

ii. K = [4.57x 10-4 M] [4.57x 10-4 M] / [0.01 M] = 2.09 x 10-5

iii. K = [1.82 x 10-4 M] [1.82 x 10-4 M] / [0.001M] = 3.31 x 10-5

Percent Dissociation = (concentration of dissociated species) / (original concentration of acid) x 100

i. % dissociation = 0.0012 / 0.1 x 100 = 1.2%

ii. % dissociation = 4.57x 10-4 / 0.01 x 100 = 4.57%

iii. % dissociation = 1.82 x 10-4 / 0.001 X 100 = 18.2%