A. One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum b

Question

A. One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum buffer capacity (equal concentrations of acetic acid and sodium acetate), and another contains 10.0 mL of pure water. Calculate the hydronium ion concentration and the pH after the addition of 0.25 mL of

0.10 M HCl to each one. What accounts for the difference in the hydronium ion concentrations? Explain this based on equilibrium concepts; in other words, saying that “one solution is a buffer” is not sufficient.

Ka of acetic acid = 1.8 x 10^-5

Acetic acid/sodium acetate buffer

Acetic acid .1M

Sodium acetate .1M

Explanation / Answer

first consider HCl added to pure water

we know that

moles = molarity x volume (L)

so

moles of HCl added = 0.1 x 0.25 x 10-3 = 2.5 x 10-5

now

final volume = 10 + 0.25 = 10.25 ml

now

[HCl ] = moles / volume (L)

[HCl] = 2.5 x 10-5 / 10.25 x 10-3

[HCl ] = 2.44 x 10-3

we know that

HCl is a very strong acid

so

100 % dissociation

HCl --> H+ + Cl-

so

[H+] = [HCl] dissocaited = 2.44 x 10-3

now

the hydronium ion concentration is 2.44 x 10-3

2)

now consider the buffer

the buffer contains an acid , CH3COOh and base Ch3COO-

so

when HCl is added

it reacts with the base CH3C00-

so

the reaction is

H+ + CH3C00- --> CH3COOH

now

initially

moles of CH3C00- = 0.1 x 10 x 10-3 = 1 x 10-3

moles of Ch3COOH = 0.1 x 10 x 10-3 = 1 x 10-3

now

moles of HCl added = 0.25 x 0.1 x 10-3 = 0.025 x 10-3

consider the reaction

H+ + Ch3OO- --> CH3COOH

moles of CH3C00- reacted = moles of H+ added = 0.025 x 10-3

moles ofo CH3C00- remaining = 0.975 x 10-3

moles of CH3COOH formed = moles of H+ added = 0.025 x 10-3

moles of CH3COOH finally = 1.025 x 10-3

now

pH = pKa + log [ CH3C00- / Ch3COOH]

pH = -log 1.8 x 10-5 + log [ 0.975 x 10-3 / 1.025 x 10-3 ]

pH = 4.723

now

-log [H+] = 4.723

[H+] = 1.89 x 10-5

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