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!

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.

4d^55s^1 ( third choice)