Question

A 10.0 mL sample of 0.75 M butyric acid is titrated with 0.50 M KOH. What is the ratio of conjugate base to weak acid, at the point where 10.0 mL of KOH has been added during the titration? Ka for butyric acid is 1.4 x 10-5

Answer 1

Answer:

Ratio of conjugate base to weak acid is 2.

Explanation:

Butyric acid (CH₃(CH₂)₂COOH) reacts with KOH (Acid-Base reaction) thus:

CH₃(CH₂)₂COOH + KOH → CH₃(CH₂)₂COO⁻ + K⁺ + H₂O

Where CH₃(CH₂)₂COO⁻ is the conjugate base.

10.0mL of 0.75M of butyric acid are:

0.010L ₓ (0.75mol / L) = 7.5x10⁻³ mol of butyric acid

That react with 10.0mL of 0.50M of KOH are:

0.010L ₓ (0.50mol / L) = 5.0x10⁻³ mol of KOH

That means you will produce 5.0x10⁻³ mol of CH₃(CH₂)₂COO⁻ and will remain:

7.5x10⁻³ mol of butyric acid - 5.0x10⁻³ mol = 2.5x10⁻³ mol of butyric acid

That means ratio of conjugate base to weak acid is:

5.0x10⁻³ mol / 2.5x10⁻³ mol = 2

I hope it helps!

Answer 2

Ratio of conjugate base to weak acid is 2.

Balanced chemical reaction:

Butyric acid (CH₃(CH₂)₂COOH) reacts with Potassium hydroxide KOH (Acid-Base reaction) thus:

CH₃(CH₂)₂COOH + KOH → CH₃(CH₂)₂COO⁻ + K⁺ + H₂O

where, CH₃(CH₂)₂COO⁻ is the conjugate base.

10.0mL of 0.75M of butyric acid are:

0.010L * (0.75mol / L) = 7.5 * 10⁻³ mol of butyric acid

That react with 10.0mL of 0.50M of KOH are:

0.010L * (0.50mol / L) = 5.0 * 10⁻³ mol of KOH

That means you will produce 5.0 * 10⁻³ mol of CH₃(CH₂)₂COO⁻ and will remain:

7.5 * 10⁻³ mol of butyric acid - 5.0 * 10⁻³ mol = 2.5 * 10⁻³ mol of butyric acid

Thus the ratio of conjugate base to weak acid is:

5.0 * 10⁻³ mol / 2.5 * 10⁻³ mol = 2

Find more information about Conjugate base here:

brainly.com/question/22514615