Name: 04/03/14 Gen. Chem. Il Benchmark I and orn 1. When acetic acid is dissolve
ID: 962951 • Letter: N
Question
Name: 04/03/14 Gen. Chem. Il Benchmark I and orn 1. When acetic acid is dissolved in water, it reacts to form hydronu acetate anions. This reaction will reach equilibrium, which left side of this equation. favors heavily the at a) What can be said about the ratio of acetic acid to acetate equlibrium? b) If you were to add NaOH (sodium hydroxide) to this solution, how would this affect the equilbrium and the proportion of acetic acid and acetate? c) If the concentration of acetic acid at equilbrium is determined to be 1.0 M and the concentration of acetate and hydronium is 0.010 M what is the equilibrlum constant at 27 °C. (The concentration of water is ignored in this calculationlExplanation / Answer
Answer – In this one we are given, acetic acid dissolve in water and form the acetate and hydronium ions.
a) When the acidic acid and acetate are in the equilibrium then there is no reaction forward or back word reaction and it is depending on the what is the change at the equilibrium. So at the equilibrium the ratio of the acetic acid to acetate is one.
b) When we added the NaOH at the equilibrium then there is reacts with acid and formed the more acetate and the reaction gets forwarded. There is acetate concertation increase and acetic acid concentration decease and the ratio of the acetic acid to the acetate is less than 1.
c) Given, at equilibrium, [CH3COOH] = 1.0 M , [CH3COO-] = 0.010 M
[H3O+] = 0.010 M , Ka = ?
We know equilibrium constant, Ka expression
Ka = [H3O+][CH3COO-] / [CH3COOH]
= 0.010 *0.010 / 1.0 M
= 1.0*10-4
So at the 27oC, the equilibrium constant, Ka is 1.0*10-4
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