Question

Acetic acid is a monoprotic weak acid with a pKa of 4.74. (Ka = 1.8 x 10-5)
(a) What is the pH of 5 mL of a 5.0% solution?
(b) What is the pH of the solution if you now add 45 ml of water to solution (a)?

How will the equivalence point volumes differ if you titrate the two solutions ?

Answer 1

(a)

Suppose we convert the percentage solution into morality as follows:

Suppose here we have Acetic acid 5%w /v.
This means 5 g in 100ml. So

Density of water = 1.00 g/mL, thus 5 g in 100 g water.

mass(m) of Acetic acid = 5.00 g

moles of Acetic acid =5.00 g/60.05 g/mol

= 0.083 moles.

Molarity of CH3COOH = 0.083 moles / Volume in L = 0.083 Mol / 5*10^-3 L

= 16.6 M

pKa of 4.74. (Ka = 1.8 x 10-5)

CH3COOH(aq) <---> CH3COO-(aq) + H+(aq)

CH3COOH(aq) <---> CH3COO-(aq) + H+(aq)
(16.6 - x)M.........................x M...............x M

Ka, or the equilibrium constant for CH3COOH, is 1.8 * 10^-5
The formula to derive it is:
[H+][A-] / [HA]
Where
[H+] is the concentration of H+ ions
[A-] is the concentration of the anionic component of acid dissolution (aka the conjugate base)
[HA] is the concentration of the acid to begin with

So putting it all together:
1.8 * 10^-5 = x^2 / (16.6 M)
x = 0.017

pH = -log[H+]
[H+] = x
pH = -log(0.017)
pH = 1.76

(b) What is the pH of the solution if you now add 45 ml of water to solution (a)?

M1V1= M2V2

16.6 M* 5.00 mL = M2*45 mL

M2= 1.84 M

pKa of 4.74. (Ka = 1.8 x 10-5)

CH3COOH(aq) <---> CH3COO-(aq) + H+(aq)

CH3COOH(aq) <---> CH3COO-(aq) + H+(aq)
(1.84 - x)M.........................x M...............x M

Ka, or the equilibrium constant for CH3COOH, is 1.8 * 10^-5
The formula to derive it is:
[H+][A-] / [HA]
Where
[H+] is the concentration of H+ ions
[A-] is the concentration of the anionic component of acid dissolution (aka the conjugate base)
[HA] is the concentration of the acid to begin with

So putting it all together:
1.8 * 10^-5 = x^2 / (1.84 M)
x = 5*75*10^-3

pH = -log[H+]
[H+] = x
pH = -log(5.75*10^-3)
pH = 2.24