HF + H2O <==> H3O+ + F-
By dissolving 0.03 mol HF in 1 L water , seen that [HF] at the equilibrium is 0.027 M , then :
1) If we divide 0.03 mol/ 1 L it would be 0.03 M , why that difference ?
2) How many moles HF in 1 L water is required to reach the equilibrium ?
1)If we dissolve 0.03 mols HF in 1L H2O, we have prepared a solution that is 0.03M in HF. However, since HF is a weak acid, it dissociates(partly) in solution to
HF + H2O ==> H3O^+ + F^- and the HF present as the molecule is decreased to 0.027M.
2) Any amount of HF in any amount of H2O will reach equilibrium on its own. I suspect you didn't post all of this problem.
1) The difference between 0.03 mol HF and 0.027 M [HF] can be explained by considering how concentration is determined. Concentration is defined as the amount of solute (in moles) divided by the volume of the solution (in liters), denoted as Molarity (M). In your question, you dissolved 0.03 mol of HF in 1 L of water. To calculate the concentration, you divide the number of moles of HF by the volume of the solution in liters.
So, if you divide 0.03 mol HF by 1 L, you get 0.03 M. However, in this case, we are given that [HF] at equilibrium is 0.027 M. The equilibrium concentration is typically different from the initial concentration due to the establishment of chemical equilibrium. During the reaction, some of the HF molecules will dissociate into H3O+ and F-. As a result, the concentration of HF decreases, and the concentration of H3O+ and F- increases.
2) To calculate how many moles of HF are required to reach equilibrium, we need to take the equilibrium concentration of HF and convert it back to moles.
Given that the equilibrium concentration of HF is 0.027 M and the volume is 1 L, we can use the formula:
moles = concentration × volume
moles = 0.027 M × 1 L
moles = 0.027 mol HF
Therefore, 0.027 moles of HF are required to reach equilibrium in 1 L of water.