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Chiman
Answers (1)
HF(aq)+H2O(l)H3O+ + F- 0.027 M XM XM mol of HF in equilibrium= 0.0300 -0.0270= 0.003 mol [H3O+] = [F-] = mol / V = 3 * 10-3 mol / 1.0 L = 3 * 10-3 M 3. Ka = [ H3O+] [F-] / [HF] = 3 * 10-3 X 3 * 10-3 / 2.7 * 10-1 = 3.3 * 10-4
