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

Answer :

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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.

1. The concentration of KOH is 0.02M

2. 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.

Chemical reaction:

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

Given:

Concentration of acid CA= 0.010 M

Volume of acid VA= 15.0 ml

Volume of base VB= 7.5 ml

Number of moles of acid NA= 1

Number of moles of base NB= 1

To find:

Concentration of base CB= ???

From equation of dilution:

CAVA/CBVB= NA/NB

CAVANB = CBVBNA

CB= CAVANB/VBNA

On substituting the values:

CB= 0.010 * 15.0 * 1 / 7.5 * 1

CB= 0.02 M

Therefore, the concentration of the KOH is 0.02M

2. 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.

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