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A 25.0 mL sample of 0.100 M acetic acid is titrated with a 0.125 M NaOH...

A 25.0 mL sample of 0.100 M acetic acid is titrated with a 0.125 M NaOH solution. Calculate the pH of the mixture after 10, 20, and 30 ml of NaOH have been added (Ka=1.76*10^-5)

Expert Solution

When you add NaOH to the CH3COOH solution you form CH3COONa. You then have a solution of a weak acid and a salt of this acid. This is a buffer solution . The pH of the buffer solution is calculated using the Henderson - Hasselbalch equation.
pH = pKa + log ([salt] /[acid])
pKa = -log Ka = - log (1.7*10^-5) = 4.77

number of moles is equal to Volume in litre multiply with Concentration.
a) 10 mL of NaOH was added.

Mol CH3COOH in 25mL of 0.1M solution = 0.025(0.1) = 0.0025 mol
Mol NaOH in 10mL of 0.125M solution = 0.01(0.125) = 0.00125 mol
The NaOH reacts with the CH3COOH in 1:1 ratio
Therefore you produce 0.00125 mol CH3COONa
and remaining unreacted is 0.00125 mol CH3COOH
In total volume 35mL = 0.035 L
Molarity of CH3COONa = 0.00125/0.35 = 0.036M
Molarity of CH3COOH = 0.00125 / 0.35 = 0.036M
Now use the H-H equation:
pH = pKa + log ([salt] /[acid])
pH = 4.77 + log ( (0.036/0.36)
pH = 4.77 + log 1
pH = 4.77 + 0
pH = 4.77.

b) 20 mL of NaOH was added.

The solution is now completely composed of a salt of a weak acid. The pH of this solution will be basic.

1) Calculate molarity of sodium acetate:

0.0025 mol / 0.045 L = 0.0001125 M

2) Calculate the K_b of sodium acetate:

K_w = K_aK_b

1.00 x 10^-14 = (1.76 x 10^-5 ) (x)

x = 5.68 x 10^-10

3) Calculate pH of the solution:

5.68 x 10^-10 = [(x) (x)] / 0.0001125

x = 6.39 x 10^-7 M (this is the hydroxide ion concentration)

pOH = -log(6.39 x 10^-7) = 6.19

pH = 14- 6.19 = 7.81.

c) 30 mL NaOH was added

You have added a very large excess of NaOH . All the acid has been neutralised. With this large excess of a strong base , the salt does not play any role in the pH of the final solution.
You have added 30mL of NaOH = 0.00375 mol NaOH
This has reacted with 25mL = 0.0025mol acid.
You have 0.00125 mol NaOH remaining unreacted in a final volume of 55mL
Molarity of NaOH solution: 0.00125/0.055 = 0.0227 M NaOH

In 0.0227M NaOH solution , [OH-] = 0.0227M
[H+] = 10^-14 / 0.0227
[H+] = 44(10^-13)

pH = -log 44(10^-14)
pH = 12.36.

queen_honey_blossom answered 2 years ago

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