[H3O+] = 1.9 x 10-5mol/L
[A-] = 1.9 x 10-5 mol/L
[HA] = 0.040 mol/L
pH = 4.73
Explanation
initial concentration of acid = 0.040 M
Ka = 8.8 x 10-9
ICE table | HA (aq) | H2O (l) | A- (aq) | H3O+ (aq) | |
Initial conc. | 0.040 M | - | 0 | 0 | |
Change | +x | - | +x | +x | |
Equilibrium conc. | 0.040 M - x | - | +x | +x |
Ka = [A-]eq[H3O+]eq / [HA]eq
8.8 x 10-9 = [(x) * (x)] / (0.040 M - x)
Solving for x, x = 1.9 x 10-5 M
[H3O+] = x = 1.9 x 10-5 M
pH = -log[H3O+]
pH = -log(1.9 x 10-5 M)
pH = 4.73
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