Chem 17.8 Part A: A buffer solution contains 0.371 M ammonium bromide and 0.364

Question

Chem 17.8

Part A:

A buffer solution contains 0.371 M ammonium bromide and 0.364 M ammonia If 0.0288 moles of hydroiodic acid are added to 225 mL of this buffer, what is the pH of the resulting solution? (Assume that the volume does not change upon adding hydroiodic acid) pH = A buffer solution contains 0.254 M nitrous acid and 0.478 M sodium nitrite If 0.0195 moles of hydroiodic acid are added to 150 mL of this buffer. what is the pH of the resulting solution? Assume that the volume does not change upon adding hydroiodic acid) pH =

Explanation / Answer

1) Molarity of nitrous acid (HNO2) = 0.254 M

Molarity of sodium nitrite (NaNO2) = 0.478 M

moles of hydroiodic acid (HI) 'C '= 0.0195

on addition of ’ C’ moles of acid to acidic buffer salt moles decreases and acid moles increases

so

Ka of nitrous acid (HNO2)= 4.0 x 10^-4

pKa = -log ka = -log (4.0 x 10^-4)

  pKa = 3.39

pH = pKa + log [Salt –C/acid + C]

     = 3.39 + log [0.478 –0.0195 / 0.254 + 0.0195]

     = 3.6

pH = 3.6

2) Molarity of Ammonium bromide = 0.371 M

   Molarity of Ammonia = 0.364 M

moles of hydroiodic acid added = 0.0288 moles

on addition of of ’ C’ moles of acid to basic buffer salt moles increases and base moles decreases

so

Kb for ammonia = 1.8 x 10^-5

pKb = -log kb = -log (1.8 x 10^-5)

  pKb = 4.75

pH = 14 –{ pKb + log[salt+ C/ base - C]}

     = 14 –{ 4.75 + log[ 0.371 + 0.0288 / 0.364 - 0.0288]}

     = 9.17

pH = 9.17

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