Question

What is the pH (to two decimal places) of a saturated aqueous solution of LiF (at 25 °C)? Ksp = 3.8 × 10–3; Ka(HF) = 6.67 × 10–4

Answer 1

To determine the pH of a saturated aqueous solution of LiF, we need to consider the equilibrium that occurs when LiF dissolves in water. The Ksp value for LiF is 3.8 × 10–3. We can use the Ksp value to calculate the concentrations of X, F, and OH- in the solution and then determine the pH.

Step-by-step explanation:

To determine the pH of a saturated aqueous solution of LiF, we need to consider the equilibrium that occurs when LiF dissolves in water. The equation is XF(s) + H₂O(l) ⇌ X(aq) + F¯(aq) + OH¯(aq). The Ksp value for LiF is 3.8 × 10–3, which is the equilibrium constant for the dissolution of LiF. We can use this Ksp value to calculate the concentrations of X, F, and OH- in the solution, and then determine the pH using these concentrations.

Answer 2

The pH of a saturated LiF solution can be determined using the provided Ksp and Ka values by finding the hydroxide ion concentration and then calculating the pOH, which in turn allows for the calculation of pH.

Step-by-step explanation:

The pH of a saturated aqueous solution of LiF at 25 °C can be found by considering the solubility product (Ksp) of LiF and the acidity constant (Ka) of HF. LiF dissociates in water to produce Li+ and F- ions. Fluoride ions F- can react with water to produce HF and OH-, slightly increasing the pH of the solution.

Since the Ksp of LiF is given as 3.8 × 10–3 and the Ka of HF is 6.67 × 10–4, we can set up the following equilibrium expressions:

1. LiF(s) → Li+(aq) + F-(aq), where Ksp = [Li+][F-]
2. F-(aq) + H2O(l) → HF(aq) + OH-(aq), where Ka = [HF][OH-]/[F-]

The equilibrium concentrations of these ions can be used to calculate the hydroxide ion concentration, [OH-], and then find the pOH:

pOH = -log[OH-]

Finally, the pH can be calculated using the relationship between pH and pOH, which at 25 °C is:

pH + pOH = 14

Thus, the pH of the solution can be found by subtracting the pOH from 14.