For the reaction LiOH (s)⇌ Li⁺ (aq)+ OH^- (aq),K_eq = 4.6 × 10^-³,and the equilibrium concentration for hydroxide ion is [OH^-] = 0.042 M.What is [Li⁺] at equilibrium?

Question

Quizplus · Accepted Answer

Using the equilibrium constant expression:
K_eq = [LiS_1+][OH^-]^1/[LiOH]
4.6 × 10^13 = [LiS_1+][0.042]^1/[LiOH]
Solving for [LiS_1+], we get:
[LiS_1+] = (4.6 × 10^13 × [LiOH])/0.042
Since the value of [LiOH] is not given, we cannot determine the exact numerical value of [LiS_1+], but we can see that it is directly proportional to [LiOH]. Therefore, as [OH^-] increases, [LiS_1+] also increases. The answer choice that reflects this relationship is A, 0.11 M.

For the Reaction LiOH (S)⇌ Li+ (Aq)+ OH- (Aq),Keq =

For the Reaction LiOH (S)⇌ Li⁺ (Aq)+ OH^- (Aq),K_eq =