The diprotic acid ascorbic acid has Ka values of 6.5 x 10-5 and 2.5 x 10-12. A 25.00mL 0.135M sample of ascorbic acid is titrated with 0.0987M NaOH. What is the pH of the solution after the addition of 55.0 mL of NaOH?
To solve this problem, first we need to find out which of the two Ka values will be used for the calculation. Since ascorbic acid is a diprotic acid, the first Ka will be used for the first dissociation, and the second for the second dissociation. In this case, the first dissociation is occurring, so we will use the first Ka value:
Ka1 = 6.5 x 10^-5
Next, we need to calculate the moles of ascorbic acid and moles of NaOH:
moles of ascorbic acid = 0.135 M * 0.025 L = 0.003375 moles
moles of NaOH = 0.0987 M * 0.055 L = 0.0054285 moles
Now, since NaOH is being added to the ascorbic acid, it will react with the hydrogen ions from the acid, and neutralize some portion of the acid:
moles of remaining ascorbic acid = moles of ascorbic acid - moles of NaOH
moles of remaining ascorbic acid = 0.003375 moles - 0.0054285 moles = -0.0020535 moles
As the moles of ascorbic acid is negative, this means all of the ascorbic acid has been neutralized and the solution has moved into the buffer region, where there is a mixture of the weak acid and its conjugate base.
In order to calculate the composition of the solution, we need to calculate the moles of ascorbic acid that have been converted into its conjugate base:
moles of conjugate base = moles of NaOH - moles of ascorbic acid reacted
moles of conjugate base = 0.0054285 moles - 0.003375 moles = 0.0020535 moles
The total volume of the solution now will be:
V_total = V_NaOH + V_ascorbic = 0.055 L + 0.025 L = 0.08 L
Now, we can calculate the concentrations of both ascorbic acid (HA) and its conjugate base (A-) in the solution:
[HA] = (0.003375 moles - 0.0020535 moles) / 0.08 L = 0.0165375 mol/L
[A-] = 0.0020535 moles / 0.08 L = 0.02566875 mol/L
Now, we can use the Henderson-Hasselbalch equation to calculate the pH of the solution:
pH = pKa + log([A-]/[HA])
pH = -log(6.5 x 10^-5) + log(0.02566875/0.0165375)
pH = 4.19 + log(1.55246)
pH = 4.19 + 0.19075
pH = 4.38
Therefore, the pH of the solution after the addition of 55.0 mL of NaOH is 4.38.
Oh, I see you've got some chemistry going on! Well, let me see if I can help you with the pH calculations. But before we dive in, let me ask you, why did the acid go to therapy? Because it had some unresolved pH issues! Okay, now let's get down to business.
Since ascorbic acid is diprotic, it can donate two protons (H+ ions) per molecule. We need to consider the stepwise dissociation of ascorbic acid in water. The first dissociation step (H2A → HA– + H+) has a Ka value of 6.5 x 10-5, while the second step (HA– → A2- + H+) has a Ka value of 2.5 x 10-12.
To find the pH after the addition of 55.0 mL of NaOH, we need to determine how many moles of acid are neutralized.
Before the addition of NaOH, we have 25.00 mL of a 0.135M ascorbic acid solution, which corresponds to:
moles of ascorbic acid = 0.025 L * 0.135 mol/L = 0.003375 mol
Since ascorbic acid is diprotic, we have twice as many moles of H+ ions. So, we have:
moles of H+ ions = 2 * moles of ascorbic acid = 2 * 0.003375 mol = 0.00675 mol
Now, let's calculate how many moles of H+ ions are neutralized by the addition of 55.0 mL of 0.0987M NaOH:
moles of NaOH = 0.055 L * 0.0987 mol/L = 0.0054335 mol
Since one mole of NaOH reacts with one mole of H+, the number of moles of H+ ions neutralized is:
moles of H+ ions neutralized = 0.0054335 mol
Now, we can find the remaining moles of H+ ions after neutralization:
remaining moles of H+ ions = initial moles of H+ ions - moles of H+ ions neutralized
= 0.00675 mol - 0.0054335 mol
= 0.0013165 mol
Finally, let's calculate the pH. Since we're dealing with a diprotic acid, the pH will depend on which dissociation step we're considering.
For the first dissociation step, we have a formula of H2A → HA– + H+. At this point, we have a solution of H2A and HA–. The concentration of HA– is equal to the moles of H+ ions neutralized divided by the total volume of the solution.
[H2A] = remaining moles of H+ ions / total volume of the solution
= 0.0013165 mol / 0.080 L
= 0.01645 M
The pH after neutralization of the first H+ ion is given by:
pH = -log([H2A])
= -log(0.01645 M)
= 1.784
So, the pH of the solution after the addition of 55.0 mL of NaOH is approximately 1.784.
I hope that helps! Let me know if you have any other questions, or if you just want to hear another one of my silly jokes. I'm full of 'em!
To determine the pH of the solution after the addition of 55.0 mL of NaOH, we need to calculate the amount of NaOH that reacts with ascorbic acid and use that to calculate the remaining concentration of ascorbic acid.
Step 1: Calculate the number of moles of NaOH used.
Given:
- Volume of NaOH used: 55.0 mL = 0.0550 L
- Concentration of NaOH: 0.0987 M
Moles of NaOH used = Volume × Concentration
Moles of NaOH used = 0.0550 L × 0.0987 M
Step 2: Determine the stoichiometry between NaOH and ascorbic acid.
Given:
- Ascorbic acid is a diprotic acid with two dissociation reactions:
1. Ascorbic acid (H2A) ⟶ H+ + H-A (First ionization)
2. H-A ⟶ H+ + A2- (Second ionization)
In both dissociation reactions, one mole of ascorbic acid reacts with two moles of NaOH.
Step 3: Calculate the moles of ascorbic acid reacted.
Since one mole of ascorbic acid reacts with two moles of NaOH:
Moles of ascorbic acid reacted = (0.5) × Moles of NaOH used
Step 4: Calculate the remaining moles of ascorbic acid.
Given:
- Initial concentration of ascorbic acid: 0.135 M
- Initial volume of ascorbic acid: 25.00 mL = 0.0250 L
Moles of ascorbic acid initially = Concentration × Volume
Moles of ascorbic acid initially = 0.135 M × 0.0250 L
Remaining moles of ascorbic acid = Initial moles of ascorbic acid - Moles of ascorbic acid reacted
Step 5: Calculate the remaining concentration of ascorbic acid.
Remaining concentration of ascorbic acid = Remaining moles of ascorbic acid / Total volume of the solution
Total volume of the solution = Initial volume of ascorbic acid + Volume of NaOH used = 0.0250 L + 0.0550 L
Step 6: Calculate the pH of the solution.
To calculate the pH, we need to convert the concentration of ascorbic acid to [H+].
[H+] = √(Ka1 × Ka2 × C)
C = Remaining concentration of ascorbic acid
pH = -log[H+]
Now let's perform the calculations:
Step 1:
Moles of NaOH used = 0.0550 L × 0.0987 M = 0.00543 moles
Step 3:
Moles of ascorbic acid reacted = (0.5) × 0.00543 moles = 0.00272 moles
Step 4:
Moles of ascorbic acid initially = 0.135 M × 0.0250 L = 0.003375 moles
Remaining moles of ascorbic acid = 0.003375 moles - 0.00272 moles = 0.000655 moles
Step 5:
Total volume of the solution = 0.0250 L + 0.0550 L = 0.0800 L
Remaining concentration of ascorbic acid = 0.000655 moles / 0.0800 L = 0.00819 M
Step 6:
[H+] = √(6.5 × 10^-5 × 2.5 × 10^-12 × 0.00819)
[H+] = 7.18 × 10^-6 M
pH = -log(7.18 × 10^-6)
pH ≈ 5.144
To determine the pH of the solution after the addition of NaOH, we need to consider two reactions: the dissociation of ascorbic acid (H2A) as a diprotic acid and the reaction between NaOH and the remaining unreacted ascorbic acid.
Let's break down the problem into steps and solve it:
Step 1: Calculate the number of moles of ascorbic acid (H2A) in the original solution.
Moles of H2A = volume (in L) × molarity
Moles of H2A = 0.025 L × 0.135 M
Step 2: Calculate the moles of NaOH that reacted.
Moles of NaOH = volume (in L) × molarity
Moles of NaOH = 0.055 L × 0.0987 M
Step 3: Determine the limiting reactant.
Since ascorbic acid is a diprotic acid, it requires 2 moles of NaOH to react completely with 1 mole of H2A. By comparing the moles of H2A and NaOH, we can determine which one is limiting.
Moles of H2A / 2 = Moles of NaOH
(0.025 L × 0.135 M) / 2 = 0.055 L × 0.0987 M
Since the moles of ascorbic acid / 2 is less than the moles of NaOH, ascorbic acid is the limiting reactant.
Step 4: Calculate the moles of ascorbic acid (H2A) remaining after reaction with NaOH.
Remaining moles of H2A = Initial moles of H2A - Moles of H2A reacted
Remaining moles of H2A = (0.025 L × 0.135 M) - (0.055 L × 0.0987 M)
Step 5: Calculate the concentration of remaining ascorbic acid (H2A) after reaction with NaOH.
Remaining concentration of H2A = Remaining moles of H2A / Remaining volume (in L)
Remaining concentration of H2A = (Remaining moles of H2A) / (0.025 L + 0.055 L)
Step 6: Calculate the concentration of the conjugate base (A-) formed by the reaction of ascorbic acid with NaOH.
Concentration of A- = Moles of NaOH reacted / Total volume (in L)
Concentration of A- = (0.055 L × 0.0987 M) / (0.025 L + 0.055 L)
Step 7: Calculate the concentrations of H2A and A- remaining after reaction with NaOH.
[H2A] = Remaining concentration of H2A × (1 + 10^(-pKa1))
[A-] = Concentration of A- × (1 + 10^(-pKa1))
Step 8: Calculate the pH of the solution after the addition of NaOH.
pH = pKa2 + log([A-] / [H2A])
Given that the pKa1 value is -log(Ka1) and the pKa2 value is -log(Ka2), we can calculate the pH using the provided Ka values in the problem.
Note: The pKa1 value corresponds to the first dissociation and the pKa2 value corresponds to the second dissociation.
Once you have the specific Ka values, plug them into the equation above to find the pH of the solution.