How much 10.0 m hno must be added to 1.00 l of a buffer that is 0.0100 m acetic acid and 0.100 m sodium acetate to reduce the ph to 4.95? 3.87 ml 2nd attempt 1st attempt c try ag view solution 13 of 22 questions completed samsung

Explanation:

It is given that molarity of acetic acid = 0.0100 M

Therefore, moles of acetic acid = molarity of acetic acid × volume of buffer

            Moles of acetic acid = 0.0100 M × 1.00 L

                                              = 0.0100 mol

Similarly, moles of acetate = molarity of sodium acetat × volume of buffer

                                           = 0.100 mol

When is added, it will convert acetate to acetic acid.

Hence, new moles acetic acid = (initial moles acetic acid) + (moles )

                                                = 0.0100 mol + x

New moles of sodium acetate = (initial moles acetate) - (moles )

                                        = 0.100 mol - x

According to Henderson - Hasselbalch equation,

           pH =

             pH =

           4.95 = 4.75 +

         = 4.95 - 4.75

                                            = 0.20

 = antilog (0.20)

                                           = 1.6

Hence,    x = 0.032555 mol

Therefore, moles of = 0.032555 mol

volume of =

                                =

                                 = 0.0032555 L

or,                             = 3.25           (as 1 L = 1000 mL)

Thus, we can conclude that volume of added is 3.26 mL.

i would say b

explanation:

but i think i would need more information than that

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.