Question

Calculate how to prepare 750 ml of 0.25 M sodium formate buffer at pH 4. Use your textbook to determine the molecular weight and pKa of the acid and base. Calculate the grams of sodium formate and number of milliliters of formic acid required. THEN using this stock solution, calculate and describe how you would prepare 100 ml of a 10 mM formate buffer, pH 3.5. By the way, what is the molarity of formic acid?

Answer 1

Explanation:

Formation of formate buffer will be as follows.

              Formate buffer = HCOOH + HCOONa

Molar mass of HCOOH $(M_{HCOOH}$ = 46 g/mol

Molar mass of HCOONa $(M_{HCOONa}$ = 68 g/mol

         $pK_{a}$ of HCOOH = 3.75

As,     $[HCOO^{-}]$ = [HCOOH] + [HCOONa]

                               = [HCOONa]    (here, dissociation of HCOOH is negligible)

Its preparation will be as follows.

          750 ml of 0.25 M sodium formate buffer

Since, there are many combinations of salt + acid and among those possibilities one of them is as follows.

              HCOONa = 0.25 M in 400 ml solution

              HCOOH = x molat in 350 ml solution

Therefore as,     pH = $pK_{a} + log \frac{N_{salt}V_{salt}}{N_{acid}V_{acid}}$

                          4 = 3.75 + $log \frac{0.25 \times 400}{x \times 350}$

                         x = 0.16 M

Therefore, molarity of the formic acid acid is 0.16 M.

Now, calculate the weight of sodium formate as follows.

                 $\frac{0.25 \times 400}{1000} \times 68$

                 = 6.8 g

And, number of milliliters of formic acid required is 350 ml.