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Shkiper50 [21]
3 years ago
7

Consider the weak acid ch3cooh (acetic acid). if a 0.048 m ch3cooh solution is 5.2% ionized, determine the [h3o+] concentration

at equilibrium.
Chemistry
1 answer:
Tatiana [17]3 years ago
3 0
To determine the equilibrium concentration of hydronium ions in the solution, we use the given value of the percent ionized. Percent ionized is the percent of the ions that is dissociated into the solution. It is equal to the concentration of an ionized species over the initial concentration of the compound multiplied by 100 percent. For this case, the dissociation of the weak acid has a 1 is to 1 ratio to the ionized species such that the concentration of the CH3COO- and H+ ions at equilibrium would be equal. We calculate as follows:

5.2% = 5.2 M H3O+ / 100 M CH3COOH
5.2 M H3O+ / 100 M CH3COOH = [H3O+] / 0.048 M CH3COOH
[H3O+] = 0.2496 M 
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35.0 mL of 12.0 M HCl is added to enough water to have a final volume of 1.20 L. What is the molarity of the final solution?
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<h3>Answer:</h3>

0.35 M

<h3>Explanation:</h3>

<u>We are given;</u>

  • Initial volume as 35.0 mL or 0.035 L
  • Initial molarity as 12.0 M
  • Final volume is 1.20 L

We are required to determine the final molarity of the solution;

  • Dilution involves adding solvent to a solution to make it more dilute which reduces the concentration and increases the solvent while maintaining solute constant.
  • Using dilution formula we can determine the final molarity.

M1V1 = M2V2

  • Rearranging the formula;

M2 = M1V1 ÷ V2

     = (12.0 M × 0.035 L) ÷ 1.2 L

      = 0.35 M

Thus, the final concentration of the solution is 0.35 M

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Explanation:

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