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Sagot :
Solution :
The buffer is the one that contains weak acid (the [tex]$\text{pK}_a$[/tex] is nearly equal to the required pH) and the salt of its conjugate base.
The [tex]$\text{pK}_a$[/tex] values of the given weak acids are as follows :
HCOOH : [tex]$\text{pK}_a$[/tex] = 3.744
[tex]$CH_3COOH : pK_a = 4.756$[/tex]
HCN : [tex]$\text{pK}_a$[/tex] = 9.21
Since the required pH is 5.03, the suitable buffer is the mixture of the acetic acid and the salt of its conjugate base.
Let suppose the volumes of [tex]$CH_3COOH$[/tex] and [tex]$CH_3COONa$[/tex] are x and y mL respectively.
So the total volume of the buffer is 1000 mL.
∴ x+ y = 1000 ................(1)
Writing the Henderson-Hasselbalch equation for the given buffer solution :
[tex]$pH = pK_a + \log \ \frac{[CH_3COONa]}{[CH_3COOH]}$[/tex] .............(2)
[tex]$5.03 = 4.756 + \log \ \frac{\left(\frac{y \ mL \times 0.10 \ M}{(x+y) \ mL}\right)}{\left(\frac{x \ mL \times 0.10 \ M}{(x+y) \ mL}\right)}$[/tex]
[tex]$5.03 = 4.756+ \log \frac{y}{x}$[/tex]
[tex]$\frac{y}{x}=10^{5.03-4.756}$[/tex]
y = 1.9 x
Substituting the values of y in equation (1), we get
x + 1.9 x = 1000
x = 345
Putting the value of x in (1), we get
345 + y = 1000
y = 655
Therefore the volume of [tex]$CH_3COOH$[/tex] is 345 mL and the volume of [tex]$CH_3COONa$[/tex] is 655 mL.
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