500.0 Ml of 0.160 M NaOH is added to 565 ml of 0.250 M weak acid. Ka=2.46x10^-5. What is the pH of the resulting buffer
mmols HA = 565 x 0.25M = 141.25
mmols NaOH = 500 x 0.160 = 80
........HA + OH^- ==> A^- + H2O
I......141.25..0......0.......0
add..........80.................
C.......-80.-80.......+80
E........?....0.......80........
Substitute the E line into the Henderson-Hasselbalch equation and solve for pH.
pH = pKa + log (base/acid)
To calculate the pH of the resulting buffer, we need to determine the concentrations of the weak acid and its conjugate base after the addition of NaOH.
Step 1: Calculate the moles of NaOH:
Moles of NaOH = volume (in liters) × concentration
Moles of NaOH = 0.500 L × 0.160 M
Moles of NaOH = 0.080 mol
Step 2: Calculate the moles of weak acid:
Moles of weak acid = volume (in liters) × concentration
Moles of weak acid = 0.565 L × 0.250 M
Moles of weak acid = 0.14125 mol
Step 3: Determine the limiting reagent:
Since NaOH and the weak acid undergo a 1:1 stoichiometric reaction, and the moles of NaOH (0.080 mol) are less than the moles of the weak acid (0.14125 mol), NaOH is the limiting reagent.
Step 4: Determine the excess of NaOH:
Excess of NaOH = Moles of NaOH - Moles of weak acid
Excess of NaOH = 0.080 mol - 0.14125 mol
Excess of NaOH = -0.06125 mol
Step 5: Calculate the final concentration of the weak acid:
The moles of the weak acid consumed in the reaction with NaOH is equal to the moles of the weak acid initially present minus the excess of NaOH.
Moles of weak acid consumed = Moles of weak acid - Excess of NaOH
Moles of weak acid consumed = 0.14125 mol - (-0.06125 mol)
Moles of weak acid consumed = 0.2025 mol
Volume of the resulting solution = Volume of weak acid (565 mL) + Volume of NaOH (500 mL) = 1.065 L
Final concentration of weak acid = Moles of weak acid consumed / Volume of resulting solution
Final concentration of weak acid = 0.2025 mol / 1.065 L
Final concentration of weak acid = 0.1901 M
Step 6: Calculate the final concentration of the conjugate base (NaA):
Since NaOH fully dissociates in water, the moles of NaOH consumed will be equal to the moles of NaOH converted to the conjugate base NaA.
Moles of NaA = Moles of NaOH consumed = Moles of weak acid consumed = 0.2025 mol
Final concentration of NaA = Moles of NaA / Volume of resulting solution
Final concentration of NaA = 0.2025 mol / 1.065 L
Final concentration of NaA = 0.1901 M
Step 7: Calculate the pKa (the negative logarithm of the acid dissociation constant Ka):
pKa = -log10(Ka)
pKa = -log10(2.46x10^-5)
pKa = 4.608
Step 8: Calculate the pH of the resulting buffer using the Henderson-Hasselbalch equation:
pH = pKa + log10([A-]/[HA])
pH = 4.608 + log10(0.1901 M / 0.1901 M)
pH = 4.608
Therefore, the pH of the resulting buffer is 4.608.
To find the pH of the resulting buffer, we need to determine the concentrations of the acid and its conjugate base after the addition of NaOH. Here's how you can approach the problem step by step:
1. Calculate the initial moles of the acid (weak acid) and its conjugate base (conjugate base of the weak acid) in the solution by using the given volumes and molarities.
Moles of acid (initial) = Volume of weak acid (in liters) x Molarity of weak acid
Moles of conjugate base (initial) = Volume of weak acid (in liters) x Molarity of conjugate base
volume of weak acid = 565 ml = 565/1000 L
volume of NaOH = 500 ml = 500/1000 L
Moles of acid (initial) = (565/1000) L x 0.250 M
Moles of conjugate base (initial) = (565/1000) L x 0.250 M
2. Calculate the moles of NaOH (strong base) added to the solution.
Moles of NaOH = Volume of NaOH (in liters) x Molarity of NaOH
= (500/1000) L x 0.160 M
3. Determine which component (acid or conjugate base) is in excess after the addition of NaOH. This will depend on the stoichiometry of the reaction between the weak acid and NaOH. Since NaOH is a strong base, it will react completely with the weak acid to form water and the conjugate base.
- If the moles of NaOH added is greater than the moles of the weak acid, then the conjugate base is in excess.
- If the moles of NaOH added is less than the moles of the weak acid, then the acid is in excess.
4. Calculate the final moles of the acid (weak acid) and its conjugate base using the stoichiometry of the reaction between the weak acid and NaOH.
- If the conjugate base is in excess:
Moles of conjugate base (final) = Moles of conjugate base (initial) + Moles of NaOH
Moles of acid (final) = Moles of acid (initial) - Moles of NaOH
- If the acid is in excess:
Moles of acid (final) = Moles of acid (initial) + Moles of NaOH
Moles of conjugate base (final) = Moles of conjugate base (initial) - Moles of NaOH
5. Calculate the concentrations of the acid and conjugate base in the final buffer solution by dividing the moles of each component by the total volume of the solution.
Concentration of acid = Moles of acid (final) / Total volume of the solution
Concentration of conjugate base = Moles of conjugate base (final) / Total volume of the solution
Total volume of the solution = (volume of weak acid + volume of NaOH)
6. Calculate the pH of the resulting buffer using the Henderson-Hasselbalch equation:
pH = pKa + log10 [(concentration of conjugate base) / (concentration of acid)]
pKa is the negative logarithm (base 10) of the acid dissociation constant Ka.
Plug in the values you calculated into the Henderson-Hasselbalch equation to find the pH of the resulting buffer.