"" What is the pH of a solution that is prepared by dissolving 8.52 grams of lactic acid (formula weight = 90.08 grams/mol) and 7.93 grams of sodium lactate (formula weight = 112.06 grams/mole) in water and diluting to 500.00 mL? The Ka for lactic acid is 0.000137.

The pH of given solution is 3.74.

Explanation:

The given data is as follows.

 Mass of lactic acid = 8.52 g,    Formula weight of lactic acid = 90.08 g/mol

So, number of moles of lactic acid will be calculated as follows.

   No. of moles =

                        =

                        = 0.094 moles

   Mass of sodium lactate = 7.93 g,    Formula weight of sodium lactate = 112.06 g/mol

Hence, number of moles of sodium lactate is as follows.

    No. of moles =

                        =

                         = 0.071 moles

As we know that relation between and is as follows.

                 = -log(0.000137)

                 = 3.86

Using Henderson equation, we will calculate the pH as follows.

      pH =

      pH =

           =

            = 3.86 + log (0.755)

            = 3.86 - 0.121

           = 3.74

Therefore, we can conclude that pH of given solution is 3.74.