when 0.20 mole HF is dissolved in water to a volume of 1.00 L, 5.8% of the HF dissociates to form F- (aq). What is the equilibrium constant for the reaction ?
HF(aq) + H2O (I) ----- F- (aq)+H3O+(aq)
0.2mol/L -= 0.2M
0.20 x 0.00.058 = 0.0116
..........HF ==> H^+ + F^-
initial...0.2.....0.....0
change.-0.0116...0.0116..0.0116
equil.....?........?......?
Ka = (H^+)(F^-)/(HF)
Substitute from the ICE chart above and solve for Ka.
Answer is 7.14x10^-4
To find the equilibrium constant for the reaction, you need to use the balanced equation and the concentrations of the reactants and products. Here is a step-by-step guide to find the equilibrium constant (K):
Step 1: Write the balanced equation for the reaction:
HF(aq) + H2O(l) → F-(aq) + H3O+(aq)
Step 2: Note that 5.8% of HF dissociates, which means that 5.8% of the initial moles of HF dissociates into F-(aq). We are given that the initial amount of HF is 0.20 mol.
Step 3: Calculate the initial concentration of HF:
Initial concentration of HF = (0.20 mol) / (1.00 L) = 0.20 M
Step 4: Calculate the concentration of F-(aq) at equilibrium:
Since 5.8% of HF dissociates, the concentration of F-(aq) will also be 5.8% of the initial concentration of HF.
Concentration of F-(aq) = 0.058 * 0.20 M = 0.0116 M
Step 5: Calculate the concentration of H3O+(aq) at equilibrium.
Since the reaction follows a 1:1 stoichiometry, the concentration of H3O+(aq) will also be 0.0116 M.
Step 6: Write the equilibrium expression using the concentrations of the ions:
K = [F-][H3O+] / [HF]
Step 7: Substitute the values into the equilibrium expression:
K = (0.0116)(0.0116) / (0.20)
Step 8: Calculate the equilibrium constant:
K ≈ 0.00067
Therefore, the equilibrium constant for the reaction is approximately 0.00067.
To find the equilibrium constant (K) for the given reaction, we need to use the expression for the equilibrium constant and the given information.
The expression for the equilibrium constant (K) is:
K = [F-][H3O+] / [HF][H2O]
Given that 5.8% of the HF dissociates to form F- (aq), we can calculate the concentrations of the species involved.
Initially, the concentration of HF is 0.20 mole in 1.00 L, so [HF] = 0.20 M.
Since 5.8% of the HF dissociates, the concentration of F- would be 5.8% of 0.20 M, which is (5.8/100) × 0.20 M = 0.0116 M. Therefore, [F-] = 0.0116 M.
Since the reaction involves water, we assume it stays constant, so the concentration of water, [H2O], remains 1.00 M.
Now, we need to find the concentration of [H3O+]. Since the reaction involves acid dissociation, we know that the concentration of [H3O+] would be the same as the concentration of the dissociated [HF]. Hence, [H3O+] = 0.0116 M.
Now, we can substitute the values into the expression for the equilibrium constant:
K = (0.0116)(0.0116) / (0.20)(1.00)
K = 0.00013456 / 0.20
K ≈ 0.0006728
Therefore, the equilibrium constant (K) for the given reaction is approximately 0.0006728.