Consider a weak acid-strong base titration in which 25.0 mL of 0.100 M acetic acid is titrated with 0.100 M NaOH.
a) Calculate the pH after the addition of 3.00mL of NaOH.
b) What is the pH of the solution before the addition of NaOH (pKa of acetic acid=4.75)
I got pH= 3.88 for part A
I agree with your answer of 3.88 for part A (but I don't know that you worked it correctly since you didn't show any work). Part B is done this way.
HAc = CH3COOH = acetic acid.
.................HAc ==> H^+ + Ac^-
initial...........0.1M... .0......0
change............-x......+x......+x
final............0.1-x.....x.......x
Ka = (H^+)(Ac^-)/(HAc)
Plug into Ka expession from the ICE chart I constructed above and solve for (H^+), then convert to pH.
To calculate the pH after the addition of NaOH in a weak acid-strong base titration, you need to understand the basic concept of acid-base titration and the equilibrium chemistry involved.
a) Calculate the pH after the addition of 3.00 mL of NaOH:
In this titration, you have a weak acid (acetic acid, CH3COOH) and a strong base (NaOH). The reaction between them can be represented as:
CH3COOH + NaOH → CH3COONa + H2O
The balanced equation shows that 1 mole of acetic acid reacts with 1 mole of NaOH to produce 1 mole of sodium acetate and 1 mole of water.
Given that you have 25.0 mL of 0.100 M acetic acid, you can calculate the number of moles of acetic acid present:
moles of acetic acid = volume (L) × concentration (M)
= 0.025 L × 0.100 M
= 0.0025 moles
Since the reaction is 1:1 between acetic acid and NaOH, adding 3.00 mL of 0.100 M NaOH means you are adding 0.003 moles of NaOH.
Now, you need to determine which species are present in the solution and at what concentrations. In this titration, the acetic acid partially dissociates, so you have a mixture of acetic acid (HA) and its conjugate base (A-), along with some sodium acetate.
To calculate the concentrations, use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
The equilibrium expression for the dissociation of acetic acid is:
CH3COOH ⇌ CH3COO- + H+
At this point in the titration, you have added enough NaOH to neutralize some acetic acid. As a result, some acetic acid will be converted to acetate ions, and the concentration ratio [CH3COO-]/[CH3COOH] has changed.
To determine the new concentrations, subtract the number of moles of NaOH added from the initial number of moles of acetic acid:
moles of acetic acid remaining = initial moles of acetic acid - moles of NaOH added
= 0.0025 moles - 0.003 moles
= -0.0005 moles
Since moles cannot be negative, this means that all the acetic acid has been neutralized, and you are left with an excess of NaOH. Therefore, the concentration of acetic acid is zero and the concentration of acetate ion is equal to the initial number of moles of acetic acid:
[CH3COOH] = 0 M
[CH3COO-] = 0.0025 moles / 0.025 L = 0.1 M
Now plug this information into the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
= 4.75 + log (0.1/0)
= 4.75 + log (undefined)
= 4.75 + undefined
Since the log of 0 is undefined, this means that the pH cannot be calculated because there is no acetic acid left. The pH will be determined by the excess NaOH added.
b) What is the pH of the solution before the addition of NaOH:
Before the addition of NaOH, the solution consists only of acetic acid. You can calculate the pH using the following equation:
pH = pKa + log ([A-]/[HA])
Substituting the known values:
pH = 4.75 + log (0/0.0025)
= 4.75 + log (0)
Since the log of 0 is undefined, this means that before the addition of NaOH, the pH cannot be calculated for the pure acetic acid solution.