a 0.451 L solution of 1.71 ascorbic acid (ka1=7.9E-5, ka2=1.6E-12) is titrated with 1.35 M NaOH. What is pH when 0.9540 L of NaOH is added?
To solve this question, we first need to determine the concentration of the ascorbic acid and the NaOH after the addition.
Step 1: Calculate the number of moles of ascorbic acid initially present.
Moles of ascorbic acid = Concentration × Volume
Moles of ascorbic acid = 1.71 × 0.451 (Since the solution is 0.451 L)
Moles of ascorbic acid = 0.77221 moles
Step 2: Determine the number of moles of NaOH added.
Moles of NaOH = Concentration × Volume
Moles of NaOH = 1.35 × 0.954 (Since 0.9540 L of NaOH is added)
Moles of NaOH = 1.2889 moles
Step 3: Calculate the resulting concentration of ascorbic acid and NaOH.
To find the resulting concentration, we need to consider the stoichiometry of the reaction between ascorbic acid and NaOH. Ascorbic acid contains two dissociable protons (H+), so it can react with two moles of NaOH per mole of ascorbic acid.
Since the molar ratio is 1:2 (ascorbic acid to NaOH), the moles of NaOH reacted will be twice the moles of ascorbic acid reacted.
Moles of NaOH reacted = 2 × Moles of ascorbic acid reacted
Moles of NaOH reacted = 2 × 0.77221
Moles of NaOH reacted = 1.54442 moles
Now, subtract the moles of NaOH reacted from the initial moles of NaOH to get the remaining moles of NaOH.
Remaining moles of NaOH = Initial moles of NaOH - Moles of NaOH reacted
Remaining moles of NaOH = 1.2889 - 1.54442
Remaining moles of NaOH = -0.25552 moles
Note: The negative sign indicates that the ascorbic acid is present in excess and has not been fully reacted by the NaOH.
Step 4: Calculate the concentration of the remaining NaOH.
Concentration of NaOH = Remaining moles of NaOH / Volume
Concentration of NaOH = -0.25552 / 0.9540
Concentration of NaOH = -0.2679 M
Step 5: Calculate the pOH of the resulting NaOH solution.
pOH = -log[OH-]
pOH = -log(0.2679)
pOH = 0.572
Step 6: Calculate the pH of the resulting solution.
pH + pOH = 14 (at 25°C)
pH = 14 - 0.572
pH ≈ 13.428
Therefore, the pH of the solution when 0.9540 L of NaOH is added is approximately 13.428.