1) Answer the following the questions regarding an acetic acid buffer solution.
ID: 1070244 • Letter: 1
Question
1) Answer the following the questions regarding an acetic acid buffer solution.
a) A solution was prepared by dissolving 0.024 moles of acetic acid (HOAc; pKa =4.8 ) in water to give 1 liter of solution. What is the pH?
b)Then 0.009 moles of sodium hydroxide (NaOH) was added in to the solution. What is the pH of the solution? (In this problem, you may ignore changes in volume due to the addition of NaOH)
c)Additionally, 0.015 moles of NaOH is then added to the solution (2). What is the pH of the solution?
Explanation / Answer
Case a:
The chemical equation and the law of mass action for this reaction are
CH3COOH <----> CH3COO- + H+
Ka = {[ CH3COO-] * [ H+] } / [CH3COOH]
[CH3COOH]
[ H+]
[ CH3COO-]
note
Initial concentration of the species
0.024
0
0
Change
-x
x
x
At equilibrium, some of the acetic acid will be ionized. We don't know right now, so let's have that be a variable; x . At equilibrium, the concentration of acetic acid will have decreased by x.
Since the stoichiometric coefficient of all species is 1, then concentrations of the ions will increase by x
At equilibrium
0.024-x
x
x
Ka = {[ CH3COO-] * [ H+] } / [CH3COOH] =(x *x)/(0.024-x)=x2/ (0.024-x) =4.8
Solving for x gives = 0.069mol/liter
pH=-log(0.069)=1.156
case b :
Determine moles of acetic acid and NaOH before mixing:
CH3COOH: 0.024 mol
NaOH: 0.009 mol
Determine moles of acetic acid and sodium acetate after mixing:
CH3COOH: 0.024-0.009= 0.015 mol
CH3COONa: 0.009 mol
The above comes from consideration of this reaction:
CH3COOH + NaOH <===> CH3COONa + H2O
Use the Henderson-Hasselbalch Equation:
pH = Ka + log (base/acid)
pH = 4.8 + log [(0.009 mol/1 L) / (0.015 mol/1 L)]
pH = 4.8 + log 0.6 = 4.578
case c:
Determine moles of acetic acid and NaOH present:
CH3COOH: 0.015 mol
NaOH: 0.015 mol
Determine moles of acetic acid and sodium acetate after mixing:
CH3COOH: 0.015-0.015= 0 mol
CH3COONa: 0.015 mol
The above comes from consideration of this reaction:
CH3COOH + NaOH <===> CH3COONa + H2O
This solution is no longer a buffer. It is now the salt of a weak base and its solution will be acidic. The H-H Equation is not used to determine the pH of the solution.
[CH3COOH]
[ H+]
[ CH3COO-]
note
Initial concentration of the species
0.024
0
0
Change
-x
x
x
At equilibrium, some of the acetic acid will be ionized. We don't know right now, so let's have that be a variable; x . At equilibrium, the concentration of acetic acid will have decreased by x.
Since the stoichiometric coefficient of all species is 1, then concentrations of the ions will increase by x
At equilibrium
0.024-x
x
x
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.