What is the pH of a buffer solution that is made by adding 21.0 grams of sodium fluoride to 1.00 Liter of 0.15M hydrofluoric acid? (see the Table above for the Ka of hydrofluoric acid; calculate the pH to 2 decimal places) hint: you will have to find the molarity of NaF in this solution.
Acid Dissociation Constant:
HF 6.6 x 10-4
What is the pH of a buffer solution that is made by adding 21.0 grams of sodium fluoride to 1.00 Liter of 0.15M hydrofluoric acid? (see the Table above for the Ka of hydrofluoric acid; calculate the pH to 2 decimal places) hint: you will have to find the molarity of NaF in this solution.
Acid Dissociation Constant:
HF 6.6 x 10-4
To find the pH of the buffer solution, we need to understand how a buffer works and how to calculate the molarity of sodium fluoride (NaF) in the solution.
A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added to it. In this case, we have a buffer solution made by adding sodium fluoride (NaF) to hydrofluoric acid (HF).
The molarity of hydrofluoric acid is given as 0.15M. We are asked to calculate the molarity of sodium fluoride (NaF) in the solution. To do this, we need to convert the given mass of NaF into moles, and then divide it by the volume of the solution.
The molar mass of sodium fluoride (NaF) is 41.99 g/mol. Therefore, the number of moles of NaF can be calculated as follows:
moles of NaF = mass of NaF / molar mass of NaF
moles of NaF = 21.0 g / 41.99 g/mol
moles of NaF ≈ 0.501 mol
Now, since the volume of the solution is given as 1.00 liter, the molarity of NaF can be calculated as follows:
Molarity of NaF = moles of NaF / volume of solution
Molarity of NaF = 0.501 mol / 1.00 L
Molarity of NaF ≈ 0.501 M
Next, we need to determine the pH of the buffer solution. The pH of a buffer solution is determined by an equilibrium reaction that takes place between the weak acid (HF) and its conjugate base (F-).
The acid dissociation constant (Ka) for hydrofluoric acid (HF) is given as 6.6 x 10^-4. Ka represents the equilibrium constant for the dissociation of HF into H+ and F- ions.
The expression for Ka is as follows:
Ka = [H+][F-] / [HF]
Since the concentration of H+ in the buffer solution is equivalent to the concentration of HF (0.15M), and the concentration of F- in the buffer solution is equivalent to the concentration of NaF (0.501M), we can substitute these values into the equation:
6.6 x 10^-4 = [H+][0.501] / [0.15]
Rearranging the equation to solve for [H+]:
[H+] = (6.6 x 10^-4)(0.15) / 0.501
[H+] ≈ 1.968 x 10^-4 M
Finally, to find the pH of the buffer solution, we need to take the negative logarithm (base 10) of the concentration of H+:
pH = -log[H+]
pH = -log(1.968 x 10^-4)
pH ≈ 3.71
Therefore, the pH of the buffer solution is approximately 3.71.