What is the PH of the solution after 25.00 ml of 0.250 M KOH is added to 75.00 ml of 120 M Acetic acid?
To determine the pH of the solution after mixing the two solutions, we need to calculate the concentration of the resulting solution of acetic acid and its conjugate base, acetate ion. Here's how you can do it:
1. Begin by calculating the moles of KOH added:
Moles of KOH = concentration of KOH × volume of KOH solution
= 0.250 M × 0.02500 L
= 0.00625 moles
2. Calculate the moles of acetic acid present initially in the solution:
Moles of acetic acid = concentration of acetic acid × volume of acetic acid solution
= 0.120 M × 0.07500 L
= 0.009 moles
3. Since KOH is a strong base, it reacts completely with acetic acid to form water and a salt, potassium acetate (CH3COOK). The moles of acetic acid remaining after the reaction will be the initial moles minus the moles of KOH added:
Moles of acetic acid remaining = moles of acetic acid initial - moles of KOH added
= 0.009 moles - 0.00625 moles
= 0.00275 moles
4. The final volume of the solution will be the sum of the two initial volumes: 0.02500 L + 0.07500 L = 0.100 L.
5. Now, calculate the concentration of acetic acid and acetate ion in the final solution:
Concentration of acetic acid = moles of acetic acid remaining / final volume in liters
= 0.00275 moles / 0.100 L
= 0.0275 M
Concentration of acetate ion (CH3COO-) = moles of KOH added / final volume in liters
= 0.00625 moles / 0.100 L
= 0.0625 M
6. Based on the equilibrium equation of acetic acid (CH3COOH) and its conjugate base (CH3COO-),
CH3COOH(aq) ⇌ H+(aq) + CH3COO-(aq)
Since acetic acid is a weak acid, we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log10 ([A-] / [HA])
The pKa of acetic acid is 4.75.
7. Plug in the values into the Henderson-Hasselbalch equation:
pH = 4.75 + log10 (0.0625 M / 0.0275 M)
8. Calculate the ratio of acetate ion concentration to acetic acid concentration:
Ratio = 0.0625 M / 0.0275 M
= 2.27
9. Take the logarithm of the ratio using a base 10 logarithm function:
log10 (2.27) ≈ 0.357
10. Add the pKa value to the logarithm:
pH ≈ 4.75 + 0.357
≈ 5.11
Therefore, the pH of the solution after adding 25.00 ml of 0.250 M KOH to 75.00 ml of 0.120 M acetic acid is approximately 5.11.