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?
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.
Ribonuclease (RNase) usually have the highest activity at pH=5. If we plan to prepare a buffer...
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