A weak acid with a Ka of 1.8 × 10–5 is titrated with a strong base. During the titration,12.5 mL of 0.10 M NaOH is added to 50.0 mL of 0.100 M acetic acid. What is the pH after the addition of the NaOH?
A.
7.22
B.
5.13
C.
4.26
D.
8.59
I know without any addition, the pH is 2.87. I don't know what it would be after addition of base. Thank you so much!
Is the answer 4.26?
The answer is 4.26
ROBERT
This is a buffer solution problem. Let's let HAc be acetic acid. Then
millimoles HAc initially = mL x M = 50 mL x 0.1 M = 5
millimoles NaOH added = 12.5 mL x 0.1 M = 1.25
.....................HAc + NaOH ==> NaAc + H2O
initial..............5...........0..............0..............0
added......................1.25..................................
change........-1.25...-1.25.............+1.25......1.25
equilibrium...3.75.......0..................1.25......1.25
Plug the equilibrium line into the Henderson-Hasselbalch buffer equation and solve for pH.
The H-H equation is pH = pKa + log (base concn/acid concn). The base is the salt NaAc and the acid is HAc remaining. For pKa find pKa = -log Ka.
Post your work if you get stuck.
To determine the pH after addition of NaOH, you will need to calculate the moles of acid and base initially, and then determine the excess base remaining after neutralization. Finally, you can use these calculations to find the concentration of the remaining acid, and consequently, the pH of the solution.
Step 1: Calculate moles of acetic acid
Molarity of acetic acid = 0.100 M
Volume of acetic acid = 50.0 mL = 0.0500 L
Moles of acetic acid = Molarity × Volume = 0.100 M × 0.0500 L = 0.00500 moles
Step 2: Calculate moles of NaOH
Molarity of NaOH = 0.10 M
Volume of NaOH = 12.5 mL = 0.0125 L
Moles of NaOH = Molarity × Volume = 0.10 M × 0.0125 L = 0.00125 moles
Step 3: Determine the limiting reagent
The reaction between acetic acid (CH3COOH) and NaOH (NaOH) results in the formation of water (H2O) and sodium acetate (CH3COONa) according to the balanced equation:
CH3COOH + NaOH → CH3COONa + H2O
In this case, the stoichiometric ratio between acetic acid and NaOH is 1:1. Therefore, the limiting reagent will be the one with fewer moles. Since acetic acid has 0.00500 moles and NaOH has 0.00125 moles, NaOH is the limiting reagent.
Step 4: Calculate the remaining moles of NaOH
Since NaOH is completely consumed during the reaction, there will be zero moles remaining.
Step 5: Determine the concentration of the remaining acetic acid
To calculate the concentration of the acetic acid, we need to determine its volume after the neutralization. It can be calculated using the initial volume and the volume of NaOH added.
Volume of acetic acid remaining = Initial volume - Volume of NaOH added
= 50.0 mL - 12.5 mL
= 37.5 mL = 0.0375 L
Concentration of acetic acid = Moles of acetic acid / Volume of acetic acid remaining
= 0.00500 moles / 0.0375 L
= 0.133 M
Step 6: Calculate the pH
The pH of the weak acid can be determined using the dissociation constant (Ka) of acetic acid. The equation for Ka is given by:
Ka = [H+][A-] / [HA]
Since acetic acid is a weak acid, it partially dissociates into its conjugate base, acetate ion (A-) and hydrogen ion (H+).
[H+] = [A-] = x (change in concentration)
[HA] = initial concentration - x
Ka = (x)(x) / (0.133 - x)
Since the value of x is expected to be very small, we can approximate (0.133 - x) as 0.133. Thus, the equation becomes:
1.8 × 10–5 = x^2 / 0.133
Rearrange the equation:
x^2 = (1.8 × 10–5) × 0.133
x^2 ≈ 2.394 × 10–6
x ≈ 0.00155 M
To convert this concentration to pH, we use the equation:
pH = -log[H+]
pH = -log(0.00155)
pH ≈ 2.81
Since we are measuring the pH after the addition of NaOH, the solution becomes basic. In basic solutions, we can determine the pOH using the equation:
pOH = 14 - pH
pOH = 14 - 2.81
pOH ≈ 11.19
Finally, we can calculate the pH using the equation:
pH = 14 - pOH
pH = 14 - 11.19
pH ≈ 2.81
Therefore, the pH after the addition of NaOH is approximately 2.81.
Since none of the answer choices match the calculated pH of 2.81, there may have been an error in the calculations or a limitation in the options provided. Please double-check the calculations and consider alternative answer choices if applicable.