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)
Using Henderson-Hasselbalch:
pH = pKa + log [HA] / [A-]
so
pH = - log (1.7•10-5) + log (9.13/5.63)
pH = 4.76 + 0.209
pH = 4.96
To calculate the pH of the buffer solution after the addition of NaOH, we will use the Henderson-Hasselbalch equation, which is given by:
pH = pKa + log([A-]/[HA])
Where:
- pH is the desired pH of the buffer solution
- pKa is the acid dissociation constant of the weak acid (acetic acid)
- [A-] is the concentration of the conjugate base (acetate ion)
- [HA] is the concentration of the weak acid (acetic acid)
First, we need to determine the moles of acetic acid and sodium acetate in the initial buffer solution. We can calculate this using the formula:
moles = concentration (M) * volume (L)
Moles of acetic acid = 0.122 M * 0.0521 L = 0.0063642 mol
Moles of sodium acetate = 0.182 M * 0.0461 L = 0.0084022 mol
Next, we need to determine the new concentrations of acetic acid and acetate ion after adding NaOH. This will involve taking into account the stoichiometry of the reaction between acetic acid and NaOH.
The balanced equation for the reaction between acetic acid and NaOH is:
CH3COOH + NaOH → CH3COONa + H2O
From the balanced equation, we can see that one mole of acetic acid reacts with one mole of NaOH to form one mole of sodium acetate and one mole of water.
Since we are adding 5.82 mL of 0.125 M NaOH, the moles of NaOH added can be calculated as follows:
moles of NaOH added = 0.125 M * 0.00582 L = 0.0007275 mol
Due to the 1:1 stoichiometry between acetic acid and NaOH, the moles of acetic acid remaining after the reaction will be:
moles of acetic acid remaining = moles of acetic acid - moles of NaOH added = 0.0063642 mol - 0.0007275 mol = 0.0056367 mol
Since the volume of the final solution is the sum of the initial volumes plus the volume of NaOH added, the total volume is:
Total volume = 52.1 mL + 46.1 mL + 5.82 mL = 104.02 mL = 0.10402 L
Now, we can calculate the concentrations of acetic acid and acetate ion after the addition of NaOH:
Concentration of acetic acid = moles of acetic acid remaining / total volume = 0.0056367 mol / 0.10402 L ≈ 0.0542 M
Concentration of acetate ion = moles of sodium acetate / total volume = 0.0084022 mol / 0.10402 L ≈ 0.0807 M
Finally, we can plug these values into the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([A-]/[HA])
pH = pKa + log(0.0807 M / 0.0542 M)
The pKa value for acetic acid is approximately 4.74. Substituting the values:
pH ≈ 4.74 + log(0.0807 M / 0.0542 M)
pH ≈ 4.74 + log(1.488)
Evaluating the logarithm:
pH ≈ 4.74 + 0.1727
pH ≈ 4.9127
Therefore, the pH of the buffer solution after the addition of 5.82 mL of 0.125 M NaOH is approximately 4.9127.