Acetic acid (HOAc) is a weak acid with a Ka= 1.8 X 10^-5 and sodium hydroxide (NaOH) is a strong base. What is the pH of a solution made by mixing 75.0 ml of a 0.160 M solution NaOH and 25.0 ml of a 0.640 M solution of HOAc?
I got a pH of 5.22, but I am not sure if this is right. Please let me know... I have a big exam tomorrow and want to make sure I am doing these correctly. Thanks.
I have 5.217 which rounds to 5.22. Good work. I have
pH = 4.74 + log (12/4)
You may not have used the same units I did for base/acid but you should have taken the log of 3 no matter what units were used.
Net Ionic Reaction:
HC2H3O2(aq) + OH^-(aq) --> C2H3O2^-(aq) + H2O
Initial moles of oH- = (0.075L)(0.160mol/L) = 0.012 mol OH- or NaOH
Initial Moles HOAc- = (0.025L)(0.640mol/L) = 0.016 mol HOAc
Final Moles of C2H3O2^- = moles OH- (limiting reagent)= 0.012
Final moles of HOAc = 0.016 - 0.012 = 0.004
pH = pKa + log([C2H3O2^-] / [HOAc])
pKa = 4.74, [C2H3O2^-]=0.012, [HOAc]= 0.004
You decide if your answer is correct
To determine the pH of the solution, we need to find the concentration of H+ ions in the solution.
First, we can calculate the moles of NaOH and HOAc used. Moles = Molarity x Volume (in L):
Moles of NaOH = (0.160 mol/L) x (0.075 L) = 0.012 mol
Moles of HOAc = (0.640 mol/L) x (0.025 L) = 0.016 mol
Since NaOH is a strong base, it completely dissociates into Na+ and OH- ions. Therefore, the concentration of OH- ions in the solution is the same as the concentration of NaOH:
Concentration of OH- = (0.012 mol) / (0.100 L) = 0.12 M
Now, let's consider the equilibrium dissociation of HOAc:
HOAc + H2O ⇌ H3O+ + OAc-
The equilibrium constant (Ka) for this reaction is given as 1.8 x 10^-5. Since the concentration of water remains essentially unchanged, we can assume it is constant and omit it from the equilibrium expression.
Ka = [H3O+][OAc-] / [HOAc]
We know the concentration of HOAc is 0.016 mol / 0.100 L = 0.16 M.
Let x represent the change in concentration of H3O+ and OAc- ions. Thus, [H3O+] = x and [OAc-] = x.
Now we can substitute these values into the equilibrium expression and solve for x:
1.8 x 10^-5 = (x)(x) / 0.16
Simplifying the equation:
1.8 x 10^-5 = x^2 / 0.16
Multiply both sides by 0.16:
0.16 x (1.8 x 10^-5) = x^2
2.88 x 10^-6 = x^2
Taking the square root of both sides, we get:
x ≈ 5.37 x 10^-3
This concentration of H3O+ ions corresponds to the pH value of the solution.
pH = -log[H3O+] = -log(5.37 x 10^-3) = 2.27
Therefore, the pH of the solution made by mixing 75.0 mL of 0.160 M NaOH and 25.0 mL of 0.640 M HOAc is approximately 2.27.
It seems like your calculation of pH as 5.22 is incorrect. The correct answer is pH = 2.27. I recommend double-checking your calculations to determine where the error occurred.