Question

Determine the pH of the following solutions

a) You dissolve 2.75 g of acetic acid with enough water in a 500.0 mL volumetric flask. Ka of acetic acid = 1.76 x 10−5

b) You dissolve 3.20 g of sodium acetate with enough water in a 500.0 mL volumetric flask.

c) You mix all of Solution A and B from above together into a larger beaker

d) You add 3.50 mL of 1.0 M HCl into Solution C (buffer)

Answer 1

a)

Moles of CH3COOH = 2.75g/60.06g/mol = 0.04579mol

molarity of CH3COOH = ( 0.04579mol/500ml) × 1000ml = 0.09158M

CH3COOH <-------> CH3COO^- + H⁺

Ka = [CH3COOH][H⁺]/[CH3COOH] = 1.76×10^-5

at equilibrium

[CH3COOH] = 0.09158 - x

[CH3COO^-] = x

[H⁺] = x

so,

x²/( 0.09158 - x) = 1.76 ×10^-5

solving for x

x = 0.001261

[H⁺] = 0.001261M

pH = -log[H⁺]

pH = -log(0.001261M)

pH = 2.90

b)

Moles of CH3COONa = 3.20g/82.04g/mol = 0.039005mol

[CH3COO^-] = (0.039005mol/500ml)×1000ml = 0.07801M

CH3COO^- partly hydrolysed by water

CH3COO^- + H2O -------> CH3COOH + OH^-

Kb = [CH3COOH][H⁺] /[ CH3COO^-]

Kb = Kw/Ka = 1.00 ×10^-14/1.76×10^-5 = 5.68 ×10^-10

at equilibrium ,

[CH3COO^-] = 0.07801 - x

[CH3COOH] = x

[OH^-] = x

so,

x²/( 0.07801 - x) = 5.68 ×10^-10

we can assume 0.07801 - x = 0.07801 because x is small value

x²/ 0.07801 = 5.68 ×10^-10

x² = 4.43× 10^-11

x = 6.66× 10^-6

[OH^-] = 6.66×10^-6M

pOH = -log[OH^-]

pOH = -log(6.66×10^-6)

pOH = 5.18

pH = 14 - pOH

pH = 14 - 5.18

pH = 8.82

c)

After mixing the solution

[CH3COOH] = 0.09158M/2 = 0.04579M

[CH3COO^-] = 0.07801M/2 = 0.039005M

Henderson - Haseelbalch equation is

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

pH = 4.75 + log ( 0.039005M/0.04579M)

pH = 4.75 - 0.07

pH = 4.68

d)

Initial moles of CH3COOH= 0.04579mol

Initial moles of CH3COO^- = 0.039005mol

HCl reacts with conjucate base CH3COO^-

H⁺ + CH3COO^- --------> CH3COOH   

moles of HCl added = (1.0mol/1000ml ) × 3.50ml = 0.0035mol

0.0035moles of HCl react with 0.0035moles of CH3COO^- to give 0.0035moles of CH3COOH

After addition of HCl

mumber of moles of CH3COOH = 0.04579 mol + 0.0035mol = 0.04929mol

number of moles of CH3COO^- = 0.035009mol - 0.0035mol = 0.031509mol

Total volume = 1000ml + 3.50ml = 1003.50ml

[CH3COOH] = (0.04929mol/1003.50ml)× 1000ml = 0.04912M

[CH3COO^-] = (0.031509mol/1003.50ml) ×1000ml = 0.03140M

Applying Henderson - Hasselbalch equation

pH = 4.75 + log ( 0.03140M/0.04912M)

pH = 4.75 - 0.19

pH = 4.56