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Alex
3 years ago
14

25) Base your answer on the information below and on your knowledge of chemistry. In a laboratory investigation, an HCl(aq) solu

tion with a pH value of 2 is used to determine the molarity of a KOH(aq) solution. A 7.5-milliliter sample of the KOH(aq) is exactly neutralized by 15.0 milliliters of the 0.010 M HCl(aq). During this laboratory activity, appropriate safety equipment is used and safety procedures are followed. Explain, in terms of aqueous ions, why 15.0 mL of a 1.0 M HCl(aq) solution is a better conductor of electricity than 15.0 mL of a 0.010 M HCl(aq) solution.
Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
3 0

Answer:

Concentration of KOH is 0.02M

15.0 mL of a 1.0 M HCl(aq) solution is a better conductor of electricity because it is more concentrated and has greater ions in solution.

Explanation:

Equation of the reaction;

KOH(aq) + HCl(aq) ----> KCl(aq) + H2O(l)

To determine the concentration of the KOH

Concentration of acid CA= 0.010 M

Volume of acid VA= 15.0 ml

Concentration of base CB= ???

Volume of base VB= 7.5 ml

Number of moles of acid NA= 1

Number of moles of base NB= 1

From;

CAVA/CBVB= NA/NB

CAVANB = CBVBNA

CB= CAVANB/VBNA

Substituting values;

CB= 0.010 × 15.0 ×1 / 7.5 ×1

CB= 0.02 M

Therefore, the concentration of the KOH is 0.02M

15.0mL of 1.0 M HCl(aq) is a better conductor of electricity than 15.0mL of 0.010 M HCl(aq) because 15.0mL of 1.0 M HCl(aq) is more concentrated and has more ions in solution than 15.0mL of 0.010 M. Since ions are charge carriers in solution, the more the concentration of ions in solutions, the greater the conductivity of the solution. Hence the answer.

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(a) Write the balanced neutralization reaction that occurs between H2SO4 and KOH in aqueous solution. Phases are optional. (b) S
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These are two questions and two answers

Answer:

    Question 1:

  • <u>H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)</u>

    Question 2:

  • <u>0.201 M</u>

Explanation:

<u>Question 1:</u>

The<em> neutralization</em> reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.

The products of an acid-base reaction are salt and water.

This is the sketch of such neutralization reaction:

1) <u>Word equation:</u>

  • sulfuric acid + potassium hydroxide → potassium sulfate + water

                 ↑                               ↑                              ↑                       ↑

               acid                          base                        salt                   water

<u>2) Skeleton equation (unbalanced)</u>

  • H₂SO₄ + KOH → K₂SO₄ + H₂O

<u>#) Balanced chemical equation (including phases)</u>

  • H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l) ← answer

<u>Question 2:</u>

<u>1) Mol ratio:</u>

Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:

  • 1 mol H₂SO₄ (aq) : 2 mol KOH (aq) : 1 mol K₂SO₄ (aq) : mol 2H₂O (l)

<u>2) Moles of H₂SO₄:</u>

  • V = 0.750 liter
  • M = 0.480 mol/liter
  • M = n/V ⇒ n = M×V = 0.480 mol/liter × 0.750 liter = 0.360 mol

<u>3) Moles of KOH:</u>

  • V = 0.700 liter
  • M = 0.290 mol/liter
  • M = n/V ⇒ n = M × V = 0.290 mol/liter × 0.700 liter = 0.203 mol

<u>4) Determine the limiting reagent:</u>

a) Stoichiometric ratio:

   1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH

b) Actual ratio:

   0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH

Since hte actual ratio of H₂SO₄  is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.

<u>5) Amount of H₂SO₄ that reacts:</u>

  • Since, KOH is the limiting reactant, using 0.203 mol KOH and the stoichiometryc ratio 1 mol H₂SO₄ / 2 mol KOH, you get:

         x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒

         x = 0.203 / 2 = 0.0677 mol of H₂SO₄

<u>6) Concentration of H₂SO₄ remaining:</u>

  • Initial amount - amount that reacted = 0.360 mol - 0.0677 mol = 0.292 mol

  • Total volume = 0.700 liter + 0.750 liter = 1.450 liter

  • Concetration = M

        M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer

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3 years ago
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