Worksheet: Polyprotic Acid Dissociation

In this worksheet, we will practice calculating concentrations of species for polyprotic acids, where multiple equilibria exist.

Q1:

Hydrogen telluride ( H T e ) 2 is a diprotic acid with 𝐾 a 1 = 2 . 3 Γ— 1 0 βˆ’ 3 and 𝐾 a 2 = 1 . 3 Γ— 1 0 βˆ’ 1 2 . For a 0.1 M solution of H T e 2 , approximately how many orders of magnitude lower is the change in [ H T e ] 2 – caused by the second dissociation relative to the first dissociation?

  • A3
  • B1
  • C6
  • D10
  • E15

Q2:

Nicotine is a base that can accept a maximum of two protons, with 𝐾 a values of 7 . 0 Γ— 1 0 βˆ’ 7 and 1 . 4 Γ— 1 0 βˆ’ 1 1 . The initial concentration of nicotine in an aqueous solution is 0.050 M.

Calculate [ C H N H ] 1 0 1 4 2 + .

  • A 9 . 9 Γ— 1 0 βˆ’ 5 M
  • B 6 . 0 Γ— 1 0 βˆ’ 5 M
  • C 3 . 5 Γ— 1 0 βˆ’ 4 M
  • D 1 . 9 Γ— 1 0 βˆ’ 4 M
  • E 1 . 9 Γ— 1 0 βˆ’ 6 M

Calculate [ C H N H ] 1 0 1 4 2 2 2 + .

  • A 1 . 4 Γ— 1 0 βˆ’ 1 1 M
  • B 4 . 5 Γ— 1 0 βˆ’ 4 M
  • C 1 . 5 Γ— 1 0 βˆ’ 4 M
  • D 5 . 7 Γ— 1 0 βˆ’ 6 M
  • E 8 . 0 Γ— 1 0 βˆ’ 5 M

Q3:

Soda water (carbonated water) is a solution of carbon dioxide in water. The solution is acidic because C O 2 reacts with water to form carbonic acid, H C O 2 3 :

In a saturated solution of C O 2 , [ H C O ] 2 3 is initially 0.033 M before dissociation takes place.

Calculate [ H ] + at equilibrium.

  • A 8 . 5 Γ— 1 0 βˆ’ 5 M
  • B 1 . 4 Γ— 1 0 βˆ’ 8 M
  • C 6 . 1 Γ— 1 0 βˆ’ 9 M
  • D 1 . 2 Γ— 1 0 βˆ’ 4 M
  • E 6 . 2 Γ— 1 0 βˆ’ 4 M

Calculate [ C O ] 3 2 – at equilibrium.

  • A 5 . 6 Γ— 1 0 βˆ’ 1 1 M
  • B 5 . 1 Γ— 1 0 βˆ’ 1 5 M
  • C 6 . 0 Γ— 1 0 βˆ’ 8 M
  • D 1 . 2 Γ— 1 0 βˆ’ 4 M
  • E 5 . 6 Γ— 1 0 βˆ’ 1 2 M

Q4:

Calculate the value of [ C H ( C O ) ] 6 4 2 2 2 – in a 0.010 M solution of phthalic acid, C H ( C O H ) 6 4 2 2 .

  • A 3 . 6 Γ— 1 0 βˆ’ 6 M
  • B 2 . 8 Γ— 1 0 βˆ’ 3 M
  • C 7 . 2 Γ— 1 0 βˆ’ 2 M
  • D 3 . 9 Γ— 1 0 βˆ’ 6 M
  • E 1 . 1 Γ— 1 0 βˆ’ 3 M

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