HF hydrolyzes according to the following equation: HF(aq)+H2O(l)<-> H3O+(aq)+F-(aq)
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?
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
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To answer these questions, we need to apply the concepts of equilibrium and the ionization of HF.
a.) To determine how many moles of HF ionize per liter of water to reach equilibrium, we subtract the remaining moles of HF at equilibrium from the initial moles of HF. In this case, the initial moles of HF is 0.0300 mol, and the remaining moles of HF at equilibrium is 0.0270 mol.
0.0300 mol HF - 0.0270 mol HF = 0.00300 mol HF
Therefore, 0.00300 mol of HF ionizes per liter of water to reach equilibrium.
b.) To find the concentrations of [F-] and [H3O+] at equilibrium, we can use the equation for the hydrolysis of HF and the stoichiometry of the reaction.
The equation is: HF(aq) + H2O(l) <-> H3O+(aq) + F-(aq)
From the balanced equation, we see that 1 mol of HF ionizes to form 1 mol of H3O+ and 1 mol of F-. Therefore, when the remaining [HF] is 0.0270 M, both [F-] and [H3O+] will also be 0.0270 M.
So, [F-] = 0.0270 M and [H3O+] = 0.0270 M.
c.) The value of Ka (acid dissociation constant) for HF can be calculated using the equilibrium concentrations of the ions [H3O+], [F-], and [HF].
The equation for the hydrolysis of HF is: HF(aq) + H2O(l) <-> H3O+(aq) + F-(aq)
The equilibrium expression is: Ka = [H3O+][F-] / [HF]
From the previous calculations, we know that [H3O+] = 0.0270 M, [F-] = 0.0270 M, and [HF] = 0.0270 M.
Substituting these values into the equilibrium expression:
Ka = (0.0270 M)(0.0270 M) / 0.0270 M
Ka = 0.0270 M
Therefore, the value of Ka for HF is 0.0270.