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.