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

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!

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

lol for the pointsss