Question

Ribonuclease (RNase) usually have the highest activity at pH=5. If we plan to prepare a buffer for RNase with 1 mol/L, 20mL CH₃COOH and 1mol/L NaOH, how much NaOH solution is supposed to be added in CH₃COOH solution?

Answer 1

1L CH3COOH contains 1mole

Thus,

1 L: 1 mole CH3COOH

20 mL = 0.002L

0.002L : 0.002 mole

Thus, concentration of CH3COOH = 0.002mole/L = 0.002M

Ka of CH3COOH = 1.8*10-5

Now,

pH = pKa + log ([Conjugate Base]/[Acid])

Thus,

5 = -log (1.8*10-5) + log ([Conjugate Base]/[Acid])

5= 4.74 +log ([Conjugate Base]/0.002)

0.26 = log ([Conjugate Base]/0.002)

[Conjugate Base]/0.002 =100.26

[Conjugate Base] =1.82*0.002

[Conjugate Base] = 3.64*10-3 mole/L

Hence, the concentration of NaOH needed for the required pH buffer= 3.64*10-3 M

Now,

1mole NaOH is present in 1L

1L : 1 mole NaOH

0.00364 L : 0.00364 mole

Now, 0.00364 L = 3.64mL

Thus,

3.64 mL of the available NaOH must be added.