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alexandr1967 [171]
3 years ago
10

5.0 mL of 0.050 M Ba(NO3)2 M and 100.0 mL of 0.10 M KIO3 are added together in a 250 mL erlenmeyer flask. In this problem, ignor

e ionic strength effects and neglect the possible presence of any complex ions in solution. a. Will Ba(IO3)2 precipitate out of solution? Support your answer with appropriate calculations. b. What are the concentrations of Ba2+ and 103 in the solution? Report your answer using 2 significant figures.
Chemistry
1 answer:
DiKsa [7]3 years ago
7 0

Answer:

The Ba(IO₃)₂ will precipitate out in the solution.

Explanation:

The solubility of Ba(IO₃)₂ is:

Ba(IO₃)₂(s) → Ba²⁺(aq) + 2 IO₃⁻(aq)

Where ksp, is defined as:

ksp =  [Ba²⁺] [IO₃⁻]² = 1.5x10⁻⁹

If the multiplication of  [Ba²⁺] [IO₃⁻] is higher than ksp, the reaction will produce Ba(IO₃)₂, that means the solid will precipitate.

In solution, concentration of Ba²⁺ and  IO₃⁻ are:

[Ba²⁺] = 0.050M × (5mL / 105mL) = <em>0.00238M</em>

[IO₃⁻] = 0.10M × (100mL / 105mL) = <em>0.0952M</em>

Replacing in ksp formula:

[0.00238] [0.0952]² = 2.16x10⁻⁵. As the result is higher than ksp, <em>The Ba(IO₃)₂ will precipitate out in the solution.</em>

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. What mass of ammonium chloride must be added to 250. mL of water to give a solution with pH
Eduardwww [97]

The mass of ammonium chloride that must be added is : ( A ) 4.7 g

<u>Given data :</u>

Volume of water ( V )  = 250 mL = 0.25 L

pH of solution = 4.85

Kb = 1.8 * 10⁻⁵

Kw = 10⁻¹⁴

Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :

NH₄CI  +  H₂O  ⇄  NH₃ + H₃O⁺

where conc of H₃O⁺

[ H₃O⁺ ] = \sqrt{Ka.C}   and Ka = Kw / Kb

∴ Ka = 5.56 * 10⁻¹⁰

Next step : Determine the concentration of H₃O⁺  in the solution

pH = - log [ H₃O⁺ ] = 4.85

∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵

Next step : Determine the concentration of NH₄CI in the solution

C = [ H₃O⁺ ]² / Ka

  = ( 1.14125 * 10⁻⁵ )² /  5.56 * 10⁻¹⁰

  = 0.359 mol / L

Determine the number of moles of NH₄CI in the solution

n = C . V

  = 0.359 mol / L  * 0.25 L =  0.08979 mole

Final step : determine the mass of ammonium chloride that must be added to 250 mL

mass = n * molar mass

         = 0.08979 * 53.5 g/mol

         = 4.80 g  ≈ 4.7 grams

Therefore we can conclude that the mass of ammonium chloride that must be added is 4.7 g

Learn more about ammonium chloride : brainly.com/question/13050932

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