Question

A 50.0 ml sample of 0.50 M acetic acid, ch3cooh is titrated with a 0.150 M NaOH solution. calculate the ph after 25.0 ml of the base have been added (ka=1.8x10^-5)

Answer 1

moles of acetic acid = molarity x volume /1000

moles of acetic acid = 0.5 x 50 /1000

moles of acetic acid = 25 x 10-3

moles of naOh = 0.15 x 25 /1000

moles of NaOH = 3.75 x 10-3

the reaction is given by

NaOH + CH3C00H = CH3COONa + H20

moles of CH3COOH reamining [ CH3C00H ] = 25 x 10-3 - 3.75 x 10-3

[CH3COOH] = 21.25 x 10-3

moles of CH3COONa formed = moles of NaOH reacted

[CH3COONa] = 3.75 x 10-3

both form a buffer solution

pH = pKa + log [salt/acid ]

pH = -l0g Ka + log [ Ch3coona / ch3cooh]

pH = -log 1.8 x 10-5 + log [ 3.75 /21.25]

pH = 4 .0

Answer 2

The NaOH reacts with the CH3COOH to form the salt CH3COONa. You then have a buffer solution . The pH of a buffer solution is calculated using the Henderson - Hasselbalch equation.
pH = pKa + log ( [salt] /[acid])

We must calculate the 3 unknowns:
pKa = -log Ka = - log 1.8*10^-5 = 4.74

Mol of CH3COOH in 50.0mL of 0.50M solution = 50/1000*0.5 = 0.025 mol
Mol NaOH in 25.0mL of 0.150M solution = 25/1000*0.150 = 0.00375 mol

These will react to produce 0.00375 mol CH3COONa and there will be 0.025-0.00375 = 0.02125 mol CH3COOH remaining unreacted. Volume of final buffer = 75.0mL
Molarity of CH3COONa = 0.00375/0.075 = 0.05M
Molarity of CH3COOH = 0.02125/0.075 = 0.283M
Now substitute into the H-H equation:
pH = 4.74 + log (0.05/0.283)
pH = 4.74 + log 0.176
pH = 4.74 + (-0.75)
pH = 3.99