Question

You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p????a=4.20) and 0.240 M sodium benzoate. How many milliliters of each solution should be mixed to prepare this buffer?

Answer 1

Explanation:

As it is given that $pK_{a}$ = 4.20 and pH is 4.0.

Now, we use Henderson-Hasslbach equation as follows.

         pH = $pK_{a} + log \frac{[salt]}{acid}$

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

         pH = $pK_{a} + log \frac{[\text{sodium benzoate}]}{\text{[benzoic acid]}}$    ........ (1)

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

        pH = $pK_{a} + log \frac{[\text{sodium benzoate}]}{\text{[benzoic acid]}}$    

        4.0 = $4.20 + log \frac{[{C_{6}H_{5}COONa}]}{[C_{6}H_{5}COOH]}$

     $log \frac{[{C_{6}H_{5}COONa}]}{[C_{6}H_{5}COOH]}$ = 4.0 - 4.20

                  = - 0.20

  $\frac{[{C_{6}H_{5}COONa}]}{[C_{6}H_{5}COOH]}$ = $10^{-0.20}$

                  = 0.613

or,         $[C_{6}H_{5}COONa]$ = $0.613 \times [C_{6}H_{5}COOH]$  ............. (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.

            $y \times 0.240 M$ = $0.613 \times x \times 0.1 M$

                            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.