What would the pH of the solution be if 25.0 mL of 0.250 M NaOH were added?

Question

A solution is prepared by dissolving 15.0 g of pure HC₂H₃O₂ and 25.0 g of NaC₂H₃O₂ in 775 mL of solution (the final volume).

The pH of the solution was found to be 4.83. What would the pH of the solution be if 25.0 mL of 0.250 M NaOH were added?

My original answer was pH= 4.85, but that was found to be incorrect.

Answer 1

The original solution of HC₂H₃O₂ and NaC₂H₃O₂ represent a buffer solution because it contains a weak acid (HC₂H₃O₂) and the conjugate base of that acid (NaC₂H₃O₂, or simply C₂H₃O₂^-).

The first thing we will do is calculate the molar concentration of each ingredient.

molar mass HC₂H₃O₂ (acetic acid) = 60.05 g/mole

molar mass NaC₂H₃O₂ (sodium acetate) = 82.03 g/mol

Total volume = 775 mls x 1 L/1000 mls = 0.775 L

[HC₂H₃O₂]= 15.0 g x 1 mol/60.05 = 0.2498 mols / 0.775 L = 0.3223 M

NaC₂H₃O₂ = 25.0 g x 1 mol/82.03 g = 0.3048 mols / 0.775 L = 0.3933 M

From the Henderson Hasselbalch equation, we have...

pH = pKa + log [ NaC₂H₃O₂] / [HC₂H₃O₂] and since we are given the pH, we can calculate the pKa

4.83 = pKa + log (0.3933/0.3223)

4.83 = pKa + 0.08646

pKa = 4.744

Now we can address the addition of 25.0 ml of 0.250 M NaOH.

moles of OH- added = 25.0 ml x 1 L/1000 mls x 0.250 mol/L = 0.00625 moles OH- added

This will react with the acid (HC₂H₃O₂) to produce the conjugate base (C₂H₃O₂^-)

HC₂H₃O₂ + OH^- ===> C₂H₃O₂^- + H₂O

0.2498mol...0.00625.......0.3048 mol..............Initial

-0.00625....-0.00625.......+0.00625................Change

0.2436............0................0.3111...................Equilibrium

The final volume after addition will be...

775 mls + 25.0 mls = 800 mls = 0.800 L

Final concentrations will be...

[HC₂H₃O₂] = 0.2436 mol/0.800 L = 0.3045 M

[C₂H₃O₂^-] = 0.3111 mo/0.800 L = 0.3889 M

Plugging into the Henderson Hasselbalch equation and solving for pH, we have...

pH = 4.744 + log (0.3889/0.3045)

pH = 4.744 + 0.106

pH = 4.850

So, I get the same answer as you, but since all the values given in the question contain 3 significant figures, perhaps they marked your answer as incorrect since a pH of 4.85 has only 2 sig.figs (the 8 and the 5) because you have taken a log, so only the numbers after the decimal would be significant. That's my only guess as to why they considered your answer to be incorrect.