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WITCHER [35]
3 years ago
13

How many total moles of ions are released when the following sample dissolves completely in water?

Chemistry
1 answer:
pogonyaev3 years ago
8 0

Answer:

                      8.55 × 10²⁴ Ions

Explanation:

                    Ammonium Chloride is an ionic compound which contains a monatomic anion (Cl⁻ ; Chloride) and a polyatomic cation (NH₄⁺ ; Ammonium).

Hence, when added in water Ammonium Chloride ionizes as;

                                      NH₄Cl   →    NH₄⁺  +  Cl⁻

Hence, we can say that it produces two ions when dissolved in water.

Also,

We know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of ions  contained by 7.1 moles of NH₄Cl, we will use following relation to first calculate the number of molecules as;

          Moles  =  Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Molecules,

          Number of Molecules  =  Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting values,

          Number of Molecules  =  7.1 mol × 6.022 × 10²³

          Number of Molecules  =  4.27 × 10²⁴ Molecules

So,

As,

                         1 Molecule of NH₄Cl contained  =  2 Ions

So,

              4.27 × 10²⁴ Molecules of NH₄Cl will contain  =  X ions

Solving for X,

                     X =  2 Ions × 4.27 × 10²⁴ Molecules / 1 Molecule

                    X =  8.55 × 10²⁴ Ions

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Determine the percent ionization of a 0.225 M solution of benzoic acid. Express your answer using two significant figures.
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Answer:

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The Ka of benzoic acid, C₇H₆O₂, is 6.46x10⁻⁵, the equilibrium in water of this acid is:

C₇H₆O₂(aq) + H₂O(l) ⇄ C₇H₅O₂⁻(aq) + H₃O⁺(aq)

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<em>Where [] are concentrations in equilibrium</em>

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X = -0.0038. False solution, there is no negative concentrations.

X = 0.00378M. Right solution.

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0.00378M / 0.225M * 100 =

<h3>1.68% is ionized</h3>
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