Question

A buffer composed of 0.50 mol acetic acid (HC₂H₃O₂) and 0.50 mol sodium acetate (Na⁺C₂H₃O₂^-) is diluted to a volume of 1.0 liter. The pH of the buffer is 4.74. How many moles of NaOH must be added to the buffer to raise its pH to 5.74?

Answer 1

From the given question,

initial conc of AcOH=0.5/1=0.5 M (Ac is acyl)

initial conc of NaOAc=0.5/1=0.5 M

pH=4.75, pKa of AcOH = 4.75 (reported in literature)

Now, after addition of NaOH, the new equilibrium is like this:

  

Now, [AcOH]= 0.5 M, Total volume= 1 ltr (assumption is that the extra volume of NaOH does not change the total volume of the solution much)

Say required moles of NaOH added = x

Thus, [OH-]=x/1 = x M

Note, as the final pH(=5.74) is still acidic, there is some acid remaining in the solution after neitralization by NaOH.

Hence, I can say [AcOH] > [OH-], i.e., remaining conc if AcOH after neutralization= (0.5-x) M and

[OAc-]=[OH-]= x

Using Henderson-Hasselbalch equation:

    where HA is a weak acid and A^- is a salt of the weak acid.

5.74=4.75+log{x/(0.5-x)}

=> log{x/(0.5-x)} = 0.99

=> x/(0.5-x) = 9.77

Solving for x, x=0.4536 M

Hence, required moles of NaOH = 0.4536