A reaction mixture is in a 5.4 L flask at 25 oC initially contains only 7.64 grams of acetic acid. (CH3COOH). Set up your I.C.E. table and use it to answer the following questions. HINT: You will need to look up the Ka for this equilibrium reaction.
CH3COOH(aq) ¡ê CH3COO-(aq) + H+(aq)
a. How many moles of Acetic acid are in the solution at equilibrium?
b. What is the Kb for this reaction?
c. What is the pH of the equilibrium solution?
d. What is the pOH of the equilibrium solution?
I get tired of typing, CH3COOH we will call HAc. That is 7.64 g/molar mass HAc = 7.64/60 = 0.127 moles. You should do it more accurately than that since I estimated the molar mass of HAc.
It ionizes as
.................HAc -->H^+ + Ac^-
begin(moles)...0.127...0.....0
change............-x......+x.....+x
equilibrium...0.127-x.....x......x
Ka = (H^+)(Ac^-)/((HAc)
a. See above.
b. Did you make a typo here? If you mean Ka, look that up in your text or on-line. If you REALLY mean Kb, it is Kw/Ka.
c. Substitute the ICE values into the Ka expression and solve for x. That will be moles H^+. Convert that to M by (H^+)= moles/L [Note: I must point out here that the problem does NOT tell you the volume of the solution. It tells you that the acetic acid is in a 5.4L flask BUT it doesn't say what else is there NOR its volume. I think the intent of the problem is that the flask CONTAINS 5.4 L of solution]. After moles H^+ are converted to M, then pH =-log(H^+).
d. pH + pOH = pKw = 14. Solve for pOH.
To answer these questions, we will use the ICE table method. ICE stands for Initial, Change, and Equilibrium.
a. How many moles of Acetic acid are in the solution at equilibrium?
1. First, write the balanced chemical equation for the reaction:
CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq)
2. Set up the ICE table:
CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq)
Initial: 7.64 g 0 0
Change: -x +x +x
Equilibrium: 7.64 g-x x x
The change in concentration for acetic acid is -x because it is decreasing, while the change in concentration for the acetate ion (CH3COO-) and hydrogen ion (H+) is +x because they are increasing.
3. Convert the initial mass of acetic acid to moles:
Molar mass of acetic acid (CH3COOH) = 60.05 g/mol
Number of moles = 7.64 g / 60.05 g/mol = 0.127 moles
4. At equilibrium, the number of moles of acetic acid is 0.127 moles - x.
b. What is the Ka for this reaction?
Ka is the acid dissociation constant and can be calculated using the equilibrium concentrations of the reactants and products.
The expression for Ka is: Ka = [CH3COO-][H+]/[CH3COOH]
At equilibrium, the concentrations are given by the ICE table: [CH3COO-] = x and [H+] = x.
Therefore, Ka = x^2 / (0.127 - x)
You will need to look up the Ka value for acetic acid and solve the expression for x to find the value.
c. What is the pH of the equilibrium solution?
pH is a measure of the acidity or basicity of a solution and is defined as the negative logarithm (base 10) of the concentration of hydrogen ions (H+).
From the ICE table, we know that the concentration of H+ at equilibrium is x.
Therefore, pH = -log(x)
d. What is the pOH of the equilibrium solution?
pOH is a measure of the alkalinity of a solution and is defined as the negative logarithm (base 10) of the concentration of hydroxide ions (OH-).
Since the reaction is a weak acid dissociation, we can assume that the concentration of OH- is negligible. Therefore, we can calculate pOH using the equation: pOH = 14 - pH.
Substitute the calculated pH from part c into this equation to find the pOH.