Let's proceed step-by-step. --- ### **Step 1: Find the moles of L-Tyrosine and NaOH** #### **Moles of L-Tyrosine** - Volume = 10.00 mL = 0.01000 L - Concentration = 0.4500 M \[ \text{Moles of L-Tyrosine} = 0.4500 \, \text{mol/L} \times 0.01000 \, \text{L} = 0.004500 \, \text{mol} \] #### **Moles of NaOH** - Volume = 20.00 mL = 0.02000 L - Concentration = 0.3000 M \[ \text{Moles of NaOH} = 0.3000 \, \text{mol/L} \times 0.02000 \, \text{L} = 0.006000 \, \text{mol} \] **Explanation:** First, calculate the number of moles for both reactants. We have 0.004500 mol L-Tyrosine and 0.006000 mol NaOH. --- ### **Step 2: Determine the reaction and which species are present** L-Tyrosine has three ionizable groups (COOH, NH₃⁺, and phenol OH). In fully protonated form (at low pH), it is \( \ce{H3Y^{2+}} \). NaOH can neutralize protons from these groups, typically starting with the most acidic (COOH). #### **Reaction sequence:** - NaOH will first neutralize the carboxyl group (pKa₁ ≈ 2.2), then the amino group (pKa₂ ≈ 9.1), then the phenol group (pKa₃ ≈ 10.1). #### **Stoichiometry** Each L-Tyrosine can accept up to 3 moles of OH⁻. - Moles of NaOH added: 0.006000 mol - Moles of L-Tyrosine: 0.004500 mol After first equivalence point (neutralizing COOH): need 0.004500 mol NaOH (all L-Tyr neutralized to zwitterion). - Excess NaOH: 0.006000 mol - 0.004500 mol = 0.001500 mol So, all COOH are neutralized. The remaining 0.001500 mol NaOH starts to neutralize the NH₃⁺ group. **Explanation:** After adding NaOH, all the carboxyl (COOH) groups are neutralized (first equivalence point reached), and the excess NaOH begins to react with the amino group. We're between the first and second equivalence points. --- ### **Step 3: Calculate the pH at this stage (Buffer Region: NH₃⁺/NH₂)** At this point, the solution contains: - 0.004500 mol of L-Tyrosine in its zwitterionic form (NH₃⁺/COO⁻) - 0.001500 mol of NaOH has further deprotonated NH₃⁺ to NH₂. This creates a buffer of L-Tyrosine with NH₃⁺ (protonated amino) and NH₂ (deprotonated amino). Let’s use the **Henderson-Hasselbalch equation** for the NH₃⁺/NH₂ buffer: \[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \] - \( pK_a \) for the amino group ≈ 9.1 - Base: amount deprotonated to NH₂ = 0.001500 mol - Acid: remaining as NH₃⁺ = 0.004500 mol - 0.001500 mol = 0.003000 mol Total volume = 10.00 mL + 20.00 mL = 30.00 mL = 0.03000 L \[ \text{pH} = 9.1 + \log \left( \frac{0.001500}{0.003000} \right) \] \[ \text{pH} = 9.1 + \log(0.5) \] \[ \text{pH} = 9.1 + (-0.301) \] \[ \text{pH} = 8.8 \] **Explanation:** At this stage, the solution is a buffer of L-Tyrosine zwitterion and its deprotonated (NH₂) form. Using the Henderson-Hasselbalch equation gives pH = 8.8. --- ## **Final Answer** **The pH of the titration solution after adding 20.00 mL of 0.3000 M NaOH to 10.00 mL of 0.4500 M fully protonated L-Tyrosine is approximately:** \[ \boxed{8.8} \]

The key concept used in answering this question is the understanding of acid-base titration involving a polyprotic amino acid like L-Tyrosine, which contains multiple ionizable groups with distinct pKa values. Initially, the fully protonated form of L-Tyrosine (H₃Y²⁺) reacts with added base (NaOH), leading to stepwise deprotonation of its acidic groups—first the carboxyl (COOH), then the amino (NH₃⁺), and finally the phenolic hydroxyl. By calculating the moles of titrant and analyte, we determine which groups are deprotonated at each stage. When the amount of base added lies between the first and second equivalence points, the solution acts as a buffer composed of the protonated and deprotonated forms of the amino group. Applying the Henderson-Hasselbalch equation with the relevant pKa value allows us to estimate the pH accurately during this buffer region. This approach hinges on the principles of stoichiometry, acid-base equilibria, and buffer chemistry.