Question

If 0.80 moles of acetic acid (HC₂H₃O₂) and 0.010 moles of hydrochloric acid were added to water to make a 1.0 L of an aqueous solution, what would the hydronium ion concentration, the PH and the acetate concentration of the solution be? Ka=1.8x10^-5

Answer 1

[acetic acid] = 0.80/1
= 0.80 M
[HCl] = 0.010/1
= 0.010 M

HCl < -- > H+ + Cl-
[H+] = [HCl]
= 0.010 M
HC2H3O2 < -- > H+ + C2H3O2-
0.8 0 0 intial
0.8-x x x at equlibrium

since, x is very small
so, 0.8-x = 0.8
Kc = ([H+]*[C2H3O2-])/[HC2H3O2]
1.8*10^-5 = (x^2/0.80)
x^2 = 14.4*10^-6
x = (3.79*10^-3)
[H+] = x
= (3.79*10^-3) M

total [H+] = 0.010 + (3.79*10^-3)
= 0.010+0.00379
= 0.01379 M

pH = -log[H+]
= -log(0.01379)
= 1.86

[C2H3O2-] = x
= (3.79*10^-3) M