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 dont know

explanation:

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"