Respuesta :
Explanation:
As it is given that [tex]pK_{a}[/tex] = 4.20 and pH is 4.0.
Now, we use Henderson-Hasslbach equation as follows.
     pH = [tex]pK_{a} + log \frac{[salt]}{acid}[/tex]
So, it is given that salt is sodium benzoate and acid is benzoic acid. Hence, the given equation will become as follows.
     pH = [tex]pK_{a} + log \frac{[\text{sodium benzoate}]}{\text{[benzoic acid]}}[/tex]   ........ (1)
Now, substitute the given values into equation (1) as follows.
    pH = [tex]pK_{a} + log \frac{[\text{sodium benzoate}]}{\text{[benzoic acid]}}[/tex]  Â
    4.0 = [tex]4.20 + log \frac{[{C_{6}H_{5}COONa}]}{[C_{6}H_{5}COOH]}[/tex]
   [tex]log \frac{[{C_{6}H_{5}COONa}]}{[C_{6}H_{5}COOH]}[/tex] = 4.0 - 4.20
         = - 0.20
 [tex]\frac{[{C_{6}H_{5}COONa}]}{[C_{6}H_{5}COOH]}[/tex] = [tex]10^{-0.20}[/tex]
         = 0.613
or, Â Â Â Â [tex][C_{6}H_{5}COONa][/tex] = [tex]0.613 \times [C_{6}H_{5}COOH][/tex] Â ............. (2) Â
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, Â Â Â Â Â x + y = 100 mL Â Â ....... (3)
And, equation (2) will become as follows.
      [tex]y \times 0.240 M[/tex] = [tex]0.613 \times x \times 0.1 M[/tex]
              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.
