Question

Calculate the mass of sodium acetate trihydrate (CH3COONa●3H2O) solid and the volume of 1.00 M acetic acid (CH3COOH) required to make 250.0 mL of 0.100 M buffer, pH 5.00.

Answer 1

CH3COONa --> CH3COO- + Na+

CH3COOH is also present

This is an acidic buffer; since there is a weak acid + conjugate base:

A buffer is any type of substance that will resist pH change when H+ or OH- is added.

This is typically achieved with equilibrium equations. Both type of buffer will resist both type of additions.

When a weak acid and its conjugate base are added, they will form a buffer

The equations that explain this phenomena are given below:

The Weak acid equilibrium:

HA(aq) <-> H+(aq) + A-(aq)

Weak acid = HA(aq)

Conjugate base = A-(aq)

Neutralization of H+ ions:

A-(aq) + H+(aq) <-> HA(aq); in this case, HA is formed, H+ is neutralized as well as A-, the conjugate

Neutralization of OH- ions:

HA(aq) + OH-(aq) <-> H2O(l) + A-(aq) ; in this case; A- is formed, OH- is neutralized as well as HA.

Note that the equilibirum equation can be mathematically manipulated in order to favour the "buffer" construction

Recall that, in equilibrium

Ka = [H+][A-]/[HA]

Multiply both sides by [HA]

[HA]*Ka = [H+][A-]

take the log(X)

log([HA]*Ka) = log([H+][A-])

log can be separated

log([HA]) + log(Ka) = log([H+]) + log([A-])

note that if we use "pKx" we can get:

pKa = -log(Ka) and pH = -log([H+])

substitute

log([HA]) + log(Ka) = log([H+]) + log([A-])

log([HA]) + -pKa = -pH + log([A-])

manipulate:

pH = pKa + log([A-]) - log([HA])

join logs:

pH = pKa + log([A-]/[HA])

which is Henderson hasselbach equations.

for acetic acid, pKa = 4.75

substitute

pH = pKa + log(CH3COO- / CH3COOH)

5 = 4.75 + log ((CH3COO- / CH3COOH)

(CH3COO- / CH3COOH) = 10^(5-4.75) = 1.77827

(CH3COO- / CH3COOH) = 1.77827

we know that toal buffer concentration:

mmol = MV = 250*0.1 = 25 mmol

(CH3COO- + CH3COOH) = 25

now..

(CH3COO- / CH3COOH) = 1.77827

CH3COO- = 1.77827*CH3COOH

1.77827*CH3COOH + CH3COOH) = 25

CH3COOH = (25)/(1+1.77827) =

CH3COOH =8.998 mmol

volume --> mmol/M = 8.998 /1 = 8.998 mL of acetic acid required

mass of sat -->

CH3COO- = 1.77827*CH3COOH =1.77827* 8.998 = 16 mmol

(CH3COONa●3H2O) --> 82.0343 + 3*18 = 136.0343

mass = mmol*MW = 136.0343*16 = 2176.5488 mg = 2.18 g of (CH3COONa●3H2O)

Then..

add 2.18 g of (CH3COONa●3H2O)

8.998 mL of acetic acid required

Add water until V = 250 mL