Question

If 0.80 moles of acetic acid (HC2H3O2) and 0.010 moles of hydrochloric acid were added to...

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

Homework Answers

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

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