1. An aqueous solution contains 0.450 M hypochlorous acid . Calculate the pH of

Question

1. An aqueous solution contains 0.450 M hypochlorous acid.

Calculate the pH of the solution after the addition of 5.58×10-2 moles of potassium hydroxide to 225 mL of this solution.
(Assume that the volume does not change upon adding potassium hydroxide)

pH =

2. An aqueous solution contains 0.343 M ammonia.

Calculate the pH of the solution after the addition of 2.46×10-2 moles of hydrochloric acid to 155 mL of this solution.
(Assume that the volume does not change upon adding hydrochloric acid.)

pH =

Explanation / Answer

1)

HOCl + KOH - - - - - - > KOCl + H2O

this reaction is 1:1 molar reaction

Initial moles of HOCl = (0.450mol/1000ml)×225ml = 0.10125

No of moles of KOH added = 0.0558

moles of HOCl remaining = 0.10125 - 0.0558 = 0.04545

[HOCl] = (0.04545mol/225ml)×1000ml = 0.202M

moles of OCl- produced = 0.0558

[ OCl-] = (0.0558mol/225ml)×1000ml= 0.248M

Henderson-Hasselbalch equation is

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

pKa of HOCl = 7.53

pH = 7.53 + log(0.248M/0.202M)

pH = 7.53 + 0.09

pH = 7.62

2)

NH3 + HCl - - - - - > NH4+ + Cl-

this is 1:1 molar reaction

Initial moles of NH3 =( 0.343mol/1000ml)×155ml = 0.053165

No of moles of HCl added = 0.0246

No of moles of NH3 after addition of HCl = 0.053165 - 0.0246 = 0.028565

No of moles of NH4+ formed = 0.0246

[NH3] = (0.028565mol/155ml)×1000ml = 0.1843M

[NH4+] = (0.0246mol/155ml)×1000ml = 0.1587M

pKa of NH4+ = 9.25

pH = pKa + log(0.1587M/0.1843M)

pH = 9.25 - 0.06

pH = 9.19

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