35.0 mL sample of 0.150 M acetic acid (HC2H3O2) (Ka = 1.8 x 10-5) is titrated with 0.150 M NaOH solution. Calculate the pH after 17.5 mL volumes of base have been added
millimoles HAc = mL x M = 35.0 x 0.150 = approx 5.25
mmols NaOH = 17.5 x 0.150 = 2.625
..........HAc + NaOH ==> NaAc + H2O
I........5.25....0........0......0
add.............2.625...............
C.......-2.625..-2.625...+2.625..+2.625
E.......2.625.....0.....2.625.....2.625
Substitute the E line into the Henderson-Hasselbalch equation and solve for pH.
To calculate the pH after adding 17.5 mL volumes of NaOH, we need to determine the moles of acetic acid and NaOH involved in the reaction.
1. Calculate the moles of acetic acid (HC2H3O2):
moles = concentration (M) x volume (L)
moles = 0.150 M x 0.0350 L
moles = 0.00525 mol
2. Since acetic acid is a weak acid, it will undergo a dissociation reaction with NaOH:
HC2H3O2 + NaOH → NaC2H3O2 + H2O
This means that for every 1 mole of acetic acid, 1 mole of NaOH will react.
3. Calculate the moles of NaOH:
moles = concentration (M) x volume (L)
moles = 0.150 M x 0.0175 L
moles = 0.00263 mol
4. To calculate the concentration of acetic acid and NaOH after the reaction, we need to consider the volume change. The total volume after adding 17.5 mL of NaOH is:
total volume = initial volume + volume of NaOH added
total volume = 35 mL + 17.5 mL
total volume = 52.5 mL = 0.0525 L
5. Calculate the new concentrations of acetic acid and NaOH:
acetic acid concentration = moles of acetic acid / total volume
acetic acid concentration = 0.00525 mol / 0.0525 L
acetic acid concentration = 0.100 M
NaOH concentration = moles of NaOH / total volume
NaOH concentration = 0.00263 mol / 0.0525 L
NaOH concentration = 0.050 M
6. Use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log (concentration of conjugate base / concentration of acid)
Since acetic acid is a weak acid, its conjugate base is the acetate ion (C2H3O2-).
pKa = -log(Ka) = -log(1.8 x 10^-5) = 4.74
pH = 4.74 + log (0.050 M / 0.100 M)
pH = 4.74 + log (0.5)
pH = 4.74 - 0.30
pH = 4.44
Therefore, the pH after adding 17.5 mL volumes of NaOH is 4.44.