Question

In: Chemistry

7. (16 point) (a) The pH of a 1.00 L buffer solution containing 0.15 M HOCl...

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.

Solutions

Expert Solution

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


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