if 10.0g of NaF and 20.0 g of HF are dissolved in water to make one liter of solution, what will the pH be? For HF, Ka=6.8x10-4.
Worked the same way with the same formula as your post under Tom.
To find the pH of the solution, we need to calculate the concentration of H+ ions. Since both NaF and HF are dissolved in water, they will dissociate to release ions. NaF will dissociate into Na+ and F- ions, while HF will partially dissociate into H+ and F- ions according to its acid dissociation constant (Ka).
First, let's calculate the number of moles of NaF and HF using their given masses and molar masses. The molar mass of NaF is 41 g/mol, and the molar mass of HF is 20 g/mol (the atomic mass of F is 19 g/mol, and one hydrogen atom is 1 g/mol).
Number of moles of NaF = mass / molar mass
= 10.0 g / 41 g/mol
= 0.2439 mol
Number of moles of HF = mass / molar mass
= 20.0 g / 20 g/mol
= 1.0 mol
Next, we need to find the concentrations of Na+, F-, and H+ ions in the solution.
Since we are making a 1 liter solution, the concentration of F- ions will be equal to the number of moles of NaF divided by the volume of the solution:
Concentration of F- ions = 0.2439 mol / 1 L
= 0.2439 M
The concentration of H+ ions from HF (which partially dissociates) can be found using the expression for the acid dissociation constant (Ka):
Ka = [H+][F-] / [HF]
Given that Ka = 6.8 x 10^(-4), and we know the concentration of F- ions (0.2439 M), we can rearrange the equation to find the concentration of H+ ions:
[H+] = Ka * [HF] / [F-]
= 6.8 x 10^(-4) * 1 M / 0.2439 M
≈ 2.79 x 10^(-3) M
Now that we have the concentration of H+ ions, we can calculate the pH of the solution using the equation:
pH = -log[H+]
pH = -log(2.79 x 10^(-3))
≈ 2.555
Therefore, the pH of the solution will be approximately 2.555.