consider 1.0 L of a solution which is 0.55 M HF and 0.2 M NaF (Ka for HF = 7.2 x 10-4). Calculate the pH after 0.10 mol of HCl has been added to the original solution. Assume no volume change on addition of HCl.
Goes along with the last one i posted, i know i have to realize that the acid will go to completion and subtract it from the base but im a little lost on how to go on
0.55M HF
0.2M F^-
millimols HF = 0.55 x 1000 mL = 550 mmoles
mmols NaF = 0.2M x 1000 mL = 200 mmols.
0.1 mol HCl/1000 = 0.1M = 100 mmols in 1000 mL.
--------------------------------
..........F^- + H^+ ==> HF
I........200....0.......550
add...........100.............
C......-100...-100.......+100
E........100....0........650
pH = pKa + log (base)/(acid)
Solve for pH. I obtained approximately 2.3 but check that carefully.
To calculate the pH after adding HCl to the solution, you need to consider the reaction that occurs between HCl and HF. Here are the steps you can follow:
1. Write down the balanced equation for the reaction between HCl and HF:
HCl + HF → H2O + ClF
2. Calculate the moles of HF initially present in the solution:
Volume of solution = 1.0 L
Concentration of HF = 0.55 M
Moles of HF = concentration × volume = 0.55 M × 1.0 L = 0.55 mol
3. Calculate the moles of HCl that have been added to the solution:
Moles of HCl added = 0.10 mol (given in the question)
4. Determine which reactant is limiting:
Since the reaction between HCl and HF goes to completion, the reactant that is completely consumed is the limiting reactant. In this case, it is HF since the moles of HCl added (0.10 mol) are greater than the moles of HF initially present (0.55 mol).
5. Calculate the moles of HF remaining after the reaction with HCl:
Moles of HF remaining = Moles of HF initially present - Moles of HCl added
= 0.55 mol - 0.10 mol
= 0.45 mol
6. Calculate the concentration of HF after the reaction:
Volume of solution remains constant at 1.0 L
Concentration of HF = Moles of HF remaining / Volume of solution
= 0.45 mol / 1.0 L
= 0.45 M
7. Calculate the concentration of F- ions (formed from the reaction) after the reaction:
The stoichiometry of the reaction tells us that the concentration of F- ions will be the same as the concentration of HF remaining because one mole of each reactant forms one mole of F- ions.
Concentration of F- ions = 0.45 M
8. Calculate the concentration of H3O+ ions in the solution:
Since HF is a weak acid and undergoes partial dissociation, we need to apply the equation for the dissociation of a weak acid:
HF ⇌ H+ + F-
This equation tells us that [H3O+] = [F-] = 0.45 M (as the concentration of F- ions remaining is the same as the amount of HF remaining).
9. Calculate the pOH (power of hydroxide concentration):
Since we have the concentration of H3O+ ions, we can calculate the pOH using the formula:
pOH = -log10[H3O+]
10. Calculate the pH:
pH + pOH = 14 (at 25°C)
pH = 14 - pOH
By following these steps, you can calculate the pH after adding 0.10 mol of HCl to the initial solution.