You are asked to prepare 500. mL 0.150 M acetate buffer at pH 5.10 using only pure acetic acid ( MW = 60.05 g/mol, p K a = 4.76 ), 3.00 M NaOH , and water. How many grams of acetic acid will be needed to prepare the 500. mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution.

4.504g of acetic acid

Explanation:

The acetic acid in reaction with NaOH produce acetate ion, thus:

CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺

That means the moles of acetate buffer comes, in the first, from the acetic acid

As you need 500mL (0,500L) of a 0.150M acetate buffer, moles are:

0.500L × (0.150mol / 1L) = 0.075 moles of acetate. That is:

0.075mol = [CH₃COO⁻] + [CH₃COOH]

Thus, grams of acetic acid you need to prepare the buffer are:

0.075 moles acetic acid × (60.05g / 1mol) = 4.504g of acetic acid

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.