Question

# a. Enough of a monoprotic acid is dissolved in water to produce a 0.0163 M solution....

a. Enough of a monoprotic acid is dissolved in water to produce a 0.0163 M solution. The pH of the resulting solution is 2.51. Calculate the Ka for the acid.

Ka= ________

b. The Ka of a monoprotic weak acid is 6.61 × 10-3. What is the percent ionization of a 0.108 M solution of this acid?

Percent ionization= ___________

a) moprotic acid dissociate as

HA + H2O H3O+ + A-

Ka  = [H3O+] [ A-] / [HA]

[H3O+] = 10-pH = 0.00309 M

[H3O+] = [A-] = 0.00309 M

substitute thsse value

Ka = (0.00309) (0.00309) / 0.0163

Ka = 0.000586 = 5.86 10-4

b)

Weak acid dissociates as AH ⇌ A- + H+

Ka = [A-][H+] / [AH]

but  [A-] = [H+] = x

Ka = [x][x] / [AH]

Substitute the value in equation

6.6110-3 = [x]2/ 0.108

[x]2 = 6.6110-30.108 = 0.00071388

[x] = 0.0267185

Concentration of H+ = 0.0267 M

% dissociation = [H+] 100 / [HA]

% dissociation = 0.0267 100 / 0.108 = 24.72 %

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