A buffer solution is prepared by mixing 99.0 mL of 0.450 M HF and 55.0 mL of 1.15 M NaF. Determine the pH of the solution after the addition of 0.01 moles NaOH (assume there is no change in volume when the NaOH is added).

Question

A buffer solution is prepared by mixing 99.0 mL of 0.450 M HF and 55.0 mL of 1.15 M NaF. Determine the pH of the solution after the addition of 0.01 moles NaOH (assume there is no change in volume when the NaOH is added).

Answer 1

millimols HF = mL x M = approx 44.55

mmols NaF = approx 63.25
0.01 mols = 10 mmols NaOH added

..........HF + OH^- ==> F^- + H2O
I........44.55..0.......63.25......
add............10............
C........-10...-10......10
E.......34.55...0......73.25

Plug the E line into the Henderson-Hasselbalch equation and solve for pH.

Answer 2

To determine the pH of the buffer solution after the addition of NaOH, we need to consider the reaction that occurs between the components of the buffer solution (HF and NaF) and the NaOH.

The reaction between HF and NaOH can be represented as follows:
HF + NaOH → NaF + H2O

Since HF is a weak acid, it will partially dissociate in water. Therefore, we need to take into account the concentrations of the dissolved species at equilibrium.

Let's calculate the initial concentrations of HF and NaF in the buffer solution:

Initial moles of HF = volume (in L) × concentration (in M) = 0.099 L × 0.450 M = 0.04455 mol

Initial moles of NaF = volume (in L) × concentration (in M) = 0.055 L × 1.15 M = 0.06325 mol

Now, let's determine the moles of NaF and HF remaining in the solution after the reaction with NaOH:

Moles of NaF remaining = initial moles of NaF - moles of NaOH (added)
= 0.06325 mol - 0.01 mol = 0.05325 mol

Moles of HF remaining = initial moles of HF

Since HF is a weak acid, we can calculate its dissociation using the Henderson-Hasselbalch equation:

pH = pKa + log ([A-]/[HA])

In this case, HF is the acidic component (HA) and NaF is the conjugate base (A-). The pKa of HF is known to be 3.17.

First, let's calculate the concentrations of HF and NaF remaining in the solution:

Concentration of NaF remaining = moles of NaF remaining / total volume of the solution (in L)
= 0.05325 mol / (0.099 L + 0.055 L) = 0.3215 M

Concentration of HF remaining = moles of HF remaining / total volume of the solution (in L)
= 0.04455 mol / (0.099 L + 0.055 L) = 0.2687 M

Now, let's substitute the values into the Henderson-Hasselbalch equation to calculate the pH:

pH = 3.17 + log (0.3215 M / 0.2687 M)
= 3.17 + log(1.196)
= 3.17 + 0.0772
= 3.2472

Therefore, the pH of the solution after the addition of 0.01 moles of NaOH is approximately 3.25.