Final answer:
The pH of the solution after the addition of 620.0 mL of LiOH is approximately 3.47.
Step-by-step explanation:
To determine the pH of the solution after the addition of 620.0 mL of LiOH, we can use the concept of titration. In this case, HF is the weak acid and LiOH is the strong base. The reaction between HF and LiOH produces water and the fluoride ion (F-). The balanced equation is:
HF (aq) + LiOH (aq) → H2O(l) + LiF(aq)
Since the concentration of HF is 0.20 M and the volume of HF is 620.0 mL, we can use the equation for dilution:
M1V1 = M2V2
Substituting the given values, we get:
(0.20 M)(620.0 mL) = (M2)(620.0 mL + 620.0 mL)
Simplifying, we find that M2 = 0.10 M (the concentration of the resulting solution).
Since LiF is a strong base, it will dissociate completely in water to produce OH- ions. Therefore, the resulting solution will be basic with a pH above 7.
However, to calculate the exact pH, we need to consider the concentration of OH- ions present in the solution. We can use the equation for the dissociation of water:
2H2O (l) → H3O+ (aq) + OH- (aq)
Given that the Ka of HF is 6.8 × 10^-4, we can find the concentration of H3O+ ions using the equation:
Ka = [H3O+][F-] / [HF]
Substituting the given values, we get:
6.8 × 10^-4 = [H3O+][0.10 M] / [0.20 M]
Simplifying, we find that [H3O+] = 3.4 × 10^-4 M
To convert the concentration of H3O+ ions to pH, we can use the equation:
pH = -log[H3O+]
Substituting the given value, we get:
pH = -log(3.4 × 10^-4)
Calculating, we find that the pH of the solution after the addition of 620.0 mL of LiOH is approximately 3.47.