posted by Chemistry Problem .
A buffer is made by mixing 52.1 mL of 0.122 M acetic acid with 46.1 mL of 0.182 M sodium acetate. Calculate the pH of this solution at 25 degrees C after the addition of 5.82 mL of 0.125 M NaOH
moles acetic acid = M x L
moles acetate = M x L.
Add NaOH. mols added = M x L. NaOH reacts with acetic acid to produce more acetate at the expense of the acetic acid. Therefore, calulate moles NaOH added, subtract from acetic acid and add to acetate. Then M = moles/L, substitute base and acid into Henderson-Hasselbalch equation and solve for pH. I get something like 4.9 but that's an estimate and you need to confirm it with more accurate calculations.
Thank you Dr.Bob222!!!
A buffer is prepared by adding 300.0 mL of 2.0 M NaOH to 500.0 mL of 2.0 M CH3COOH. What is the pH of this buffer?
Say: acetic acid = HA; acetate = A-
The reaction: OH- + HA → A- + H2O
Calculate the amounts present for each species:
• 52.1mL of 0.122 M acetic acid:
52.1mL x 0.122mol/1,000mL = 0.0063562mol acetic acid (HA) = 6.36mmoles
• 46.1mL of 0.182M sodium acetate:
46.1mL x 0.182mol/1,000mL = 0.0083902mol sodium acetate (A-) = 8.40mmoles
• 5.82 mL of 0.125 M NaOH:
5.82mL x 0.125mol/1,000mL = 0.0007275mol NaOH (OH-) = 0.73mmoles
Using the ICE table:
HA OH- A-
I 6.36 0.73 8.4
C - 0.73 - 0.73 + 0.73
E 5.63 0 9.13
I: initial amount (concentration);
C: amount (concentration) at the change;
E: amount (concentration) at equilibrium (sum of I + C)
pH = pKa + log [HA] / [A-]
pH = - log (1.7•10-5) + log (9.13/5.63)
pH = 4.76 + 0.209
pH = 4.96