In: Chemistry
7. (16 point) (a) The pH of a 1.00 L buffer solution containing
0.15 M HOCl and 0.15 M NaOCl is 7.52.
HOCl (aq) + H20 (l) ---> OCl- (aq) + H3O+ (aq)
(a) Calculate the pH change when 0.40 g of NaOH is added to 1.00 L
of a buffer solution containing 0.15 M HOCl and 0.15 M NaOCl.
Neglect any volume change. Ka of HOCl is 3.0 x 10-8.
(b) Calculate the pH change when 0.200 L of 0.100 M HCl is added to
1.00 L of a buffer solution containing 0.15 M HOCl and 0.15 M
NaOCl. Assume the total volume of the solution is 1.20 L. Ka of
HOCl is 3.0 x 10-8.
Q7. (a) Ka of HOCl = 3.0 x 10-8
pKa = -log(Ka)
pKa = -log(3.0 x 10-8)
pKa = 7.52
mass of NaOH added = 0.40 g
moles of NaOH added = (mass of NaOH added) / (molar mass of NaOH)
moles of NaOH added = (0.40 g) / (40 g/mol)
moles of NaOH added = 0.010 mol
concentration of NaOH in buffer = (moles of NaOH added) / (volume of buffer)
concentration of NaOH in buffer = (0.010 mol) / (1.00 L)
concentration of NaOH in buffer = 0.010 M
NaOH is a strong base which will neutralize HOCl to NaOCl
new concentration of HOCl = (initial concentration of HOCl) - (concentration of NaOH in buffer)
new concentration of HOCl = (0.15 M) - (0.010 M)
new concentration of HOCl = 0.14 M
new concentration of NaOCl = (initial concentration of NaOCl) + (concentration of NaOH in buffer)
new concentration of NaOCl = (0.15 M) + (0.010 M)
new concentration of NaOCl = 0.16 M
According to Henderson - Hasselbalch equation
pH = pKa + log([conjugate base] / [weak acid])
pH = pKa + log([NaOCl] / [HOCl])
pH = 7.52 + log(0.16 M / 0.14 M)
pH = 7.58
(b) Concentration of HCl = 0.100 M
volume of HCl added = 0.200 L
moles of HCl added = (Concentration of HCl) * (volume of HCl added)
moles of HCl added = (0.100 M) * (0.200 L)
moles of HCl added = 0.0200 mol
initial moles of HOCl = (concentration of HOCl) * (volume of buffer)
initial moles of HOCl = (0.15 M) * (1.00 L)
initial moles of HOCl = 0.15 mol
Similarly, initial moles of NaOCl = 0.15 mol
HCl is a very strong acid which will neutralize NaOCl to HOCl
new moles HOCl = (initial moles of HOCl) + (moles of HCl added)
new moles HOCl = (0.15 mol) + (0.02 mol)
new moles HOCl = 0.17 mol
new concentration of HOCl = (new moles HOCl) / (total volume)
new concentration of HOCl = (0.17 mol) / (1.20 L)
new concentration of HOCl = 0.142 M
new moles NaOCl = (initial moles of NaOCl) - (moles of HCl added)
new moles NaOCl = (0.15 mol) - (0.02 mol)
new moles NaOCl = 0.13 mol
new concentration of NaOCl = (new moles NaOCl) / (total volume)
new concentration of NaOCl = (0.13 mol) / (1.20 L)
new concentration of NaOCl = 0.108 M
According to Henderson - Hasselbalch equation
pH = pKa + log([conjugate base] / [weak acid])
pH = pKa + log([NaOCl] / [HOCl])
pH = 7.52 + log(0.108 M / 0.142 M)
pH = 7.41