As a technician in a large pharmaceutical research firm, you need to produce 350. ml of 1.00 m potassium phosphate buffer solution of ph = 7.07. the pka of h2po4− is 7.21. you have the following supplies: 2.00 l of 1.00 m kh2po4 stock solution, 1.50 l of 1.00 m k2hpo4 stock solution, and a carboy of pure distilled h2o. how much 1.00 m kh2po4 will you need to make this solution?

You need to add 203 mL of 1,00M KH₂PO₄ and 147 mL of 1,00M K₂HPO₄.

Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸; pka=7,21

Thus, Henderson–Hasselbalch equation for 7,07 phosphate buffer is:

7,07 = 7,21 + log₁₀

0,7244 = (1)

As the buffer concentration must be 1,00 M:

1,00 = [H₂PO₄⁻] + [HPO4²] (2)

Replacing (2) in (1):

[H₂PO₄⁻] = 0,5799 M

Thus:

[HPO4²] = 0,4201 M

To obtain these concentrations you need to add:

0,5799 M × 0,350 L × = 0,203 L ≡ 203 mL of 1,00M KH₂PO₄

And:

0,4201 M × 0,350 L × = 0,203 L ≡ 147 mL of 1,00M K₂HPO₄

I hope it helps!

what is the question here. i think you messed up.

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.