Question

What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as a buffer made from 475 mL of 0.200 M benzoic acid and 25 mL of 2.00 M NaOH? Look up Ka values on the formula sheet.

Answer 1

Explanation:

The given data is as follows.

      [HCOOH] = 0.2 M,       [NaOH] = 2.0 M,

         V = 500 ml,   [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

   No. of moles of benzoic acid = Molarity × Volume

                         = $2 \times 0.475$

                         = 0.095 mol

And, moles of NaOH present in the solution will be as follows.

    No. of moles of NaOH = Molarity × Volume

                          = $2 \times 0.025$

                          = 0.05 mol

Hence, the ICE table for the chemical equation will be as follows.

         $C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O$

Initial:        0.095           0.05            0             0

Equlbm:  (0.095 - 0.05)  0            0.05

        pH = $pK_{a} + log \frac{Base}{Acid}$  

              = $4.2 + log \frac{0.05}{0.045}$

              = 4.245

For,  

         $HCOOH + NaOH \rightarrow HCOONa + H_{2}O$

Initial:       0.2x     2(0.5 - x)               0

Equlbm:   0.2x - 2(0.5 - x)                 0             2(0.5 - x)

As,

           pH = $pK_{a} + log \frac{Base}{Acid}$  

          4.245 = 3.75 + $log \frac{Base}{Acid}$

      $log \frac{Base}{Acid}$ = 0.5

    $\frac{Base}{Acid}$ = 3.162

Now,

        $\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)}$ = 3.162

               x = 0.464 L

Volume of NaOH = (0.5 - 0.464) L

                             = 0.036 L

                             = 36 ml               (as 1 L = 1000 mL)

And, volume of formic acid is 464 mL.