A 0.050 M solution of the weak acid HA has [H3O+]= 3.77 x 10…
A 0.050 M solution of the weak acid HA has [H3O+]= 3.77 x 10-4 M at equilibrium. What is the Ka for the acid?HA(aq) + H2O(l) H3O+(aq) + A-(aq) (Hint: construct the ICE table, Also note that [0.05- 3.77 x 10-4 ]= 0.05)Ka = [H3O+][A-][HA]
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