The approximate pKa values for the three dissociating groups of lysince are 2.2, 9.0, 10.5. What would the pH after addition of 10ml of 1.0M NaOH to 100ml of 0.1M lysine at pH 2.2?
To solve this problem, we need to consider the reaction between NaOH and lysine and how it affects the pH.
First, let's understand the dissociating groups of lysine and their pKa values:
1. Carboxyl group (-COOH) with a pKa of 2.2.
2. Amino group (-NH2) with a pKa of 9.0.
3. Side chain amino group (-NH3+) with a pKa of 10.5.
Now, let's break down the steps to find the final pH after the addition of NaOH:
Step 1: Calculate the initial moles of lysine.
Given that the initial volume of lysine is 100 ml and the concentration is 0.1M, we can calculate the initial number of moles of lysine using the formula:
Moles = concentration × volume
Moles of lysine = 0.1M × 0.1 L (100 mL = 0.1 L) = 0.01 moles
Step 2: Determine the number of moles of NaOH added.
Given that the volume of NaOH added is 10 ml and the concentration is 1.0M, we can calculate the number of moles using the same formula as before:
Moles of NaOH = 1.0M × 0.01 L (10 mL = 0.01 L) = 0.01 moles
Step 3: Analyze the reaction between NaOH and lysine.
NaOH reacts with the acidic protons (H+) of lysine's dissociating groups. In this case, we have two acidic protons: one from the carboxyl group and one from the amino group of lysine.
Step 4: Identify the limiting reagent.
Since the moles of NaOH and lysine are equal, neither is in excess and can be considered a limiting reagent.
Step 5: Determine the moles of acid remaining after the reaction.
We will use the Henderson-Hasselbalch equation to calculate the moles of acid remaining. The equation is:
pH = pKa + log [A-]/[HA]
Where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid. In this case, [A-] represents the remaining moles of lysine (as it loses its acidic proton).
Let's calculate the remaining moles of lysine:
- For the carboxyl group, the pKa is 2.2, and the pH is given as 2.2. Therefore, [A-]/[HA] = 1/1, and the moles of remaining lysine are 0.01 moles.
- For the amino group, the pKa is 9.0. Using the Henderson-Hasselbalch equation, we can calculate the concentration of [A-]/[HA]:
pH = 9.0 + log ([A-]/[HA])
2.2 = 9.0 + log ([A-]/[HA])
-6.8 = log ([A-]/[HA])
10^(-6.8) = [A-]/[HA]
[A-]/[HA] = 1.59 × 10^(-7)
The remaining moles of lysine are 1.59 × 10^(-7) moles.
Step 6: Calculate the final concentration of lysine.
The final volume of the solution is 100 ml + 10 ml = 110 ml. We calculated the remaining moles of lysine as 0.01 moles and 1.59 × 10^(-7) moles from the carboxyl group and amino group, respectively. We need to convert these moles to concentrations:
Final concentration of lysine = (moles of lysine / final volume of solution) × 1000
= (0.01 moles / 0.11 L) × 1000
= 0.0909 M
Step 7: Determine the final pH.
Since we have calculated the final concentration of lysine, we can use the Henderson-Hasselbalch equation to find the final pH. In this case, we will use the pKa of the side chain amino group (10.5) as it will determine the pH:
pH = pKa + log ([A-]/[HA])
pH = 10.5 + log (0.0909/1)
pH = 10.5 - log(1)
pH = 10.5
Therefore, the final pH after the addition of 10 ml of 1.0M NaOH to 100 ml of 0.1M lysine at pH 2.2 is 10.5.