HF hydrolyzes according to the following equation: HF(aq)+H2O(l)<-> H3O+(aq)+F-(aq)

Question

When 0.0300 mol of HF dissolves in 1.0 L of water, the solution quickly ionizes to reach equilibrium. At equilibrium, the remaining [HF]=0.0270 M.
a.) How many moles of HF ionize per liter of water to reach equilibrium?
0.0300 mol HF - 0.0270 mol = 0.00300 mol HF
b.) What are [F-] and [H3O+]?
c.) what is the value of Ka for HF?

Answer 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

Answer 2

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Answer 3

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Answer 4

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