Question

you have to prepare a ph 5.03 buffer, and you have the following 0.10m solutions available: hcooh, hcoona, ch3cooh, ch3coona, hcn, and nacn. How many milliliters of each solution would you use to make approximately a liter of the buffer?

Answer 1

Solution :

The buffer is the one that contains weak acid (the $\text{pK}_a$ is nearly equal to the required pH) and the salt of its conjugate base.

The $\text{pK}_a$  values of the given weak acids are as follows :

HCOOH : $\text{pK}_a$  = 3.744

$CH_3COOH : pK_a = 4.756$

HCN : $\text{pK}_a$ = 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 $CH_3COOH$ and $CH_3COONa$ 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 :

$pH = pK_a + \log \ \frac{[CH_3COONa]}{[CH_3COOH]}$   .............(2)

$5.03 = 4.756 + \log \ \frac{\left(\frac{y \ mL \times 0.10 \ M}{(x+y) \ mL}\right)}{\left(\frac{x \ mL \times 0.10 \ M}{(x+y) \ mL}\right)}$

$5.03 = 4.756+ \log \frac{y}{x}$

$\frac{y}{x}=10^{5.03-4.756}$

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 $CH_3COOH$ is 345 mL and the volume of $CH_3COONa$ is 655 mL.