You need to prepare 100.0 ml of a ph 4.00 buffer solution using 0.100 m benzoic acid (=4.20) and 0.240 m sodium benzoate. how many milliliters of each solution should be mixed to prepare this buffer?

Explanation:

As it is given that = 4.20 and pH is 4.0.

Now, we use Henderson-Hasslbach equation as follows.

         pH =

So, it is given that salt is sodium benzoate and acid is benzoic acid. Hence, the given equation will become as follows.

         pH =    ........ (1)

Now, substitute the given values into equation (1) as follows.

        pH =    

        4.0 =

      = 4.0 - 4.20

                  = - 0.20

  =

                  = 0.613

or,         =  ............. (2)  

Since, the total volume of (acid + base) is given as 100.0 mL. So, let us assume volume of acid is x and volume of base is y.

Hence,           x + y = 100 mL    ....... (3)

And, equation (2) will become as follows.

            =

                            y = 0.255x

Substituting value of y into equation (3) as follows.

                        x + 0.255x = 100 mL

                        1.255 x = 100 mL

                              x = 79.68 mL

So, value of y will be as follows.

                           y = 0.255 x

                              = 0.255 × 79.68 mL

                              = 20.31 mL

Thus, we can conclude that volume of benzoic acid is 79.68 mL and volume of sodium benzoate is 20.31 mL.

i believe the answer is three water molecules