Calculate the change in pH when 41.0 mL of a 0.540 M solution of NaOH is added to 1.00 L of a solution that is 1.00 M in sodium acetate and 1.00 M in acetic

Question

so I calculate the pH before adding NaOH which is 4.75 but I can't figure out the next step

Answer 1

1.00 M sodium acetate + 1.00 M acetic acid is a BUFFER solution.

You are correct in the initial pH because when [salt] = [acid], the pH = pKa which for acetic acid is 4.74.

Now, to find final pH, you need to find final concentrations of acetate (Ac-) and acetic acid (HAc) and plug them back into the Henderson Hasselbalch equation of pH = pKa + log [acetate]/[acetic acid]

Initial moles Ac- = 1 L x 1 mole/L = 1 mole

Initial moles HAc = 1 L x 1 mole/L = 1 mole

Reaction:  NaOH + HAc ==> NaAc + H2O  SO NOTE THE [HAc] DECREASES AND [Ac-] INCREASES

moles OH- added = 0.041 L x 0.540 mol/L = 0.0221 moles

Final moles Ac- = 1 + 0.0221 = 1.0221 moles Ac-

Final moles HAc = 1 - 0.0221 = 0.9779 moles HAc

Final volume = 1 L + 0.041 L = 1.041 L

Final [Ac-] = 1.0221 moels/1.041 L = 0.9818 M

Final [HAc] = 0.9779 moles/1.041 L = 0.9394 M

Final pH = 474 + log (0.9818/0.9394)

pH = 4.74 + 0.0192

pH = 4.76

 

If you've learned about ICE Tables, you can do it easier that way.