Ask Question
16 March, 13:32

For the reaction: MgF2 (s) ⇌ Mg2 + (aq) + 2F - (aq), Ksp = 6.4 * 10-9, the addition of 0.10 M NaF to the solution cause what effect on the solubility equilibrium? View Available Hint (s) For the reaction: MgF2 (s) ⇌ Mg2 + (aq) + 2F - (aq), Ksp = 6.4 * 10-9, the addition of 0.10 M NaF to the solution cause what effect on the solubility equilibrium? Shifts the equilibrium to the right, reduces solubility. Shifts the equilibrium to the left, reduces solubility. Shifts the equilibrium to the left, increases solubility. Shifts the equilibrium to the right, increases solubility.

+5
Answers (1)
  1. 16 March, 13:58
    0
    Shifts the equilibrium to the left. reduces solubility.

    Explanation:

    MgF2 (s) ↔ Mg2 + (aq) + 2F - (aq)

    S S 2S

    ∴ Ksp = 6.4 E-9 = [ Mg2 + ] * [ F - ]² = S * (2S) ²

    ⇒ 4S² * S = 6.4 E-9

    ⇒ 4S³ = 6.4 E-9

    ⇒ S³ = 1.6 E-9

    ⇒ S = 1.1696 E-3 M

    NaF (s) → Na + (aq) + F - (aq)

    0.10M 0.10M 0.10M

    MgF2 (s) ↔ Mg2 + (aq) + 2F - (aq)

    S' S' 2S' + 0.10

    ⇒ Ksp = 6.4 E-9 = (S') * (2S' + 0.10) ²

    If we compare the concentration (0.10 M) of the ion with Ksp (6.4 E-9); thne we can neglect S' as adding:

    ⇒ 6.4 E-9 = (S') * (0.10) ² = 0.01S'

    ⇒ S' = 6.4 E-7 M

    ∴ % S' = (6.4 E-7 / 0.1) * 100 = 6.4 E-4% <<< 5%, we can make the assumption

    We can observe that S >> S' (1.1696 E-3 M >> 6.4 E-7 M), which shows that the solubility is reduced by the efect of the common ion from the salt, which causes the equilibrium to shift to the left, precipitating part of MgF2 (s).
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “For the reaction: MgF2 (s) ⇌ Mg2 + (aq) + 2F - (aq), Ksp = 6.4 * 10-9, the addition of 0.10 M NaF to the solution cause what effect on the ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers