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.

                 

Answer 2

36 ml of NaOh and 464 ml of HCOOH would be enough to form 500 ml of a buffer with the same pH as the buffer made with benzoic acid and NaOH.

We can arrive at this answer through the following calculation:

• Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.

Amount of moles of NaOH: $2 * 0.025 = 0.05 mol$

Amount of moles of benzoic acid: $2*0.475=0.095mol$

• In this case, we can calculate the pH produced by the buffer of these two reagents, as follows:

$pH=pK{a}+log\frac{base}{acid}$

$4.2+log\frac{0.05}{0.045}=4.245$

We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows:

$pH=pK{a} +log\frac{base}{acid} \\4.245=3.75+log\frac{base}{acid} \\log\frac{base}{acid}=0.5\\\frac{base}{acid} = 3.162$

• Now we must solve the equation above. This will be done using the following values:

$\frac{2(0.5-x)}{0.2x-2(0.5-x)} =0.464L$

• With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.

NaOH volume:

$(0.5-0.464) L\\0.036L ----- 36mL$

HCOOH volume:

$500 mL-36mL = 464 mL$

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