The acetate ion,CH₃COO^-,reacts with water as a weak base: CH₃COO^-(aq)+ H₂O( )$\leftrightarrows$ CH₃COOH(aq)+ OH^-(aq),with K = 5.7 $\times$10^-10 at 25 $\circ$ C.If sodium acetate were dissolved in water to make a 0.50 M solution,what would be the resulting concentration of OH^-?

Question

Quizplus · Accepted Answer

The equilibrium equation for the reaction is: 
CH3COO- + H2O ⇌ CH3COOH + OH-
The equilibrium constant expression is: 
Kb = [CH3COOH][OH-] / [CH3COO-]
Kb = 5.7×10^-10
The initial concentration of CH3COO- is 0.50 M, and the concentration of OH- is x M. At equilibrium, the concentration of CH3COOH will also be x M (because one mole of acid is produced for each mole of base reacted). Therefore, the equilibrium constant expression can be written as:
5.7×10^-10 = (x)(x) / (0.50-x)
Solving for x gives x = 1.7×10^-5 M or 1.7×10^-9 mol/L.
Therefore, the resulting concentration of OH- is 1.7×10^-5 M or 1.7×10^-9 mol/L. 
Choice B is the only option that matches this value.

The Acetate Ion,CH3COO-,reacts with Water as a Weak Base: CH3COO-(aq) ⇆\leftrightarrows⇆

The Acetate Ion,CH₃COO^-,reacts with Water as a Weak Base: CH₃COO^-(aq) $\leftrightarrows$