We’re being asked to calculate the Ka of a 0.17 M solution of a weak acid with a pH of 5.35
Recall that weak acids partially dissociate in water and that acids donate H+ to the base (water in this case). Assuming that the acid is monoprotic, the dissociation of weak acid HA is as follows:
HA(aq) + H2O(l) ⇌ A-(aq) + H3O+(aq) ; Ka = ?
The Ka expression for HA is:
If the pH = 5.35 for a 0.17 M solution of a weak acid (HA), what is Ka for the acid?
a. 2.71x10-11 M
b. 1.17x10-10 M
c. 8.14x10-8 M
d. 6.01x10-10 M
e. 8.12x10-8 M
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