A 190 mL sample of 0.293 M NaCH3CO2(aq) is diluted to 750 mL. What is the concentration of the acetic acid at equilibrium? Ka =1.8×10−5.
NaC2H3O2 = NaAc
acetic acid = HC2H3O2 = HAc
NaAc diluted = 0.293 x (190/750) = approx 0.074 but you need to confirm this and all other calculations that follow. I estimate and round.
.......Ac^- + HOH ==> HAc + OH^-
I....0.074............0......0
C.......-x............x......x
E...0.074-x...........x......x
Kb for Ac^- = (Kw/Ka for HAc) =
(x)(x)/(0.074-x)
Solve for x = (HAc) = ?
I
To find the concentration of acetic acid at equilibrium, we need to understand the equilibrium reaction and how the dilution affects the concentration.
First, let's write the balanced equation for the reaction:
CH3CO2- (aq) + H2O (l) ⇌ CH3COOH (aq) + OH- (aq)
In this reaction, NaCH3CO2 (sodium acetate) dissociates to form CH3CO2- ions and Na+ ions. Then, CH3CO2- ions react with water to form acetic acid (CH3COOH) and hydroxide ions (OH-).
Given:
Initial volume of sodium acetate solution (V1) = 190 mL = 0.190 L
Final volume after dilution (V2) = 750 mL = 0.750 L
Initial concentration of sodium acetate (C1) = 0.293 M
Equilibrium constant (Ka) = 1.8 × 10^(-5)
To solve this problem, we'll use the following equation:
Ka = [CH3COOH] [OH-] / [CH3CO2-]
We don't have the concentration of OH-, but we know that water is neutral, so [H+] = [OH-]. Therefore, we can assume that the concentration of OH- at equilibrium is equal to the concentration of H+ (acetic acid) at equilibrium.
Let's solve the problem step by step:
Step 1: Calculate the moles of sodium acetate (CH3CO2-) initially present.
Moles of CH3CO2- = C1 * V1 = 0.293 M * 0.190 L
Step 2: Calculate the moles of sodium acetate (CH3CO2-) remaining after dilution.
Moles of CH3CO2- remaining after dilution = Moles of CH3CO2- initially present * (V1/V2)
= (0.293 M * 0.190 L) * (0.190 L / 0.750 L)
Step 3: Calculate the concentration of CH3CO2- at equilibrium.
Concentration of CH3CO2- at equilibrium = Moles of CH3CO2- remaining after dilution / V2
Step 4: Calculate the concentration of H+ (acetic acid) at equilibrium using the equilibrium constant (Ka).
Ka = [CH3COOH] * [H+] / [CH3CO2-]
[CH3COOH] = Ka * [CH3CO2-] / [H+]
Since [H+] = [CH3COOH], we can use [CH3COOH] = Ka * [CH3CO2-] / [CH3CO2-]
So, [CH3COOH] = Ka
Therefore, the concentration of acetic acid (H+) at equilibrium is equal to the equilibrium constant Ka.
Substituting the given value of Ka in the equation, the concentration of acetic acid at equilibrium is 1.8 × 10^(-5) M.