Let's solve this step by step. ### **Step 1: Identify the Main Species Present and Use the First \(pK_a\)** Since the solution is of arginine hydrochloride (H₂Arg⁺), and the concentration is 0.0126 M, we start with the fully protonated form H₂Arg⁺. - The relevant equilibria for the first deprotonation is: \[ \text{H}_2\text{Arg}^+ \leftrightharpoons \text{HArg} + \text{H}^+ \] - The first \(pK_a\) is 1.823, so this is a moderately strong acid. Calculate the initial pH by assuming full dissociation for the first \(pK_a\): \[ K_{a1} = 10^{-1.823} = 0.015 \] Let \(x\) be the amount dissociated: \[ \text{H}_2\text{Arg}^+ \rightarrow \text{HArg} + \text{H}^+ \] \[ K_{a1} = \frac{[\text{HArg}][\text{H}^+]}{[\text{H}_2\text{Arg}^+]} \] \[ 0.015 = \frac{x^2}{0.0126 - x} \] Since \(K_{a1}\) is similar in magnitude to the starting concentration, we cannot neglect \(x\): Solve for \(x\) (which is \([\text{H}^+]\)): \[ 0.015(0.0126 - x) = x^2 \] \[ 0.000189 - 0.015x = x^2 \] \[ x^2 + 0.015x - 0.000189 = 0 \] Quadratic formula: \[ x = \frac{-0.015 + \sqrt{(0.015)^2 + 4 \cdot 0.000189}}{2} \] \[ x = \frac{-0.015 + \sqrt{0.000225 + 0.000756}}{2} \] \[ x = \frac{-0.015 + \sqrt{0.000981}}{2} \] \[ x = \frac{-0.015 + 0.0313}{2} \] \[ x = \frac{0.0163}{2} = 0.00815 \] So, \([\text{H}^+] = 0.00815\) M **Explanation:** We set up the equilibrium for the first deprotonation and solved the quadratic for \([\text{H}^+]\). --- ### **Step 2: Calculate the pH** \[ \text{pH} = -\log_{10}(0.00815) = 2.09 \] **Explanation:** We use the calculated \([\text{H}^+]\) to find the pH. --- ### **Step 3: Calculate All Species' Concentrations** We already have: - \([\text{H}^+] = 0.00815\) M - \([\text{H}_2\text{Arg}^+]= 0.0126 - x = 0.00445\) M - \([\text{HArg}] = x = 0.00815\) M Now, find \([\text{Arg}^-]\) and \([\text{H}_3\text{Arg}^{2+}]\): #### a. \([\text{Arg}^-]\) Second deprotonation: \[ \text{HArg} \leftrightharpoons \text{Arg}^- + \text{H}^+ \] \[ K_{a2} = 10^{-8.991} = 1.02 \times 10^{-9} \] \[ K_{a2} = \frac{[\text{Arg}^-][\text{H}^+]}{[\text{HArg}]} \] \[ [\text{Arg}^-] = K_{a2} \cdot \frac{[\text{HArg}]}{[\text{H}^+]} = 1.02 \times 10^{-9} \cdot \frac{0.00815}{0.00815} = 1.02 \times 10^{-9} \, \text{M} \] #### b. \([\text{H}_3\text{Arg}^{2+}]\) Reverse of first deprotonation: \[ \text{H}_3\text{Arg}^{2+} \leftrightharpoons \text{H}_2\text{Arg}^+ + \text{H}^+ \] \[ K_{a1} = \frac{[\text{H}_2\text{Arg}^+][\text{H}^+]}{[\text{H}_3\text{Arg}^{2+}]} \] \[ [\text{H}_3\text{Arg}^{2+}] = \frac{[\text{H}_2\text{Arg}^+][\text{H}^+]}{K_{a1}} = \frac{0.00445 \cdot 0.00815}{0.015} \] \[ = \frac{0.0000363}{0.015} = 0.00242 \, \text{M} \] **Explanation:** We use the equilibrium expressions for the subsequent deprotonations, plugging in the known concentrations and Ka values. --- ## **Final Answers** - **pH** = **2.09** - \([\text{H}_3\text{Arg}^{2+}]\) = **0.00242 M** - \([\text{H}_2\text{Arg}^+]\) = **0.00445 M** - \([\text{HArg}]\) = **0.00815 M** - \([\text{Arg}^-]\) = **1.02 \times 10^{-9} M** Let me know if you want to see the intermediate steps for the other concentrations!

The concept used involves understanding acid-base equilibria and the pKₐ values of arginine's ionizable groups to determine the distribution of all species in solution and calculate the pH. Starting with the initial fully protonated form, H₂Arg⁺, the first step is to identify which protonation state predominates based on the pKₐ values and initial concentration. Using the equilibrium expressions for each deprotonation, the concentrations of H₃Arg²⁺, H₂Arg⁺, HArg, and Arg⁻ are calculated by applying the Henderson-Hasselbalch equation and solving the quadratic equations derived from the dissociation constants. This approach allows for the accurate determination of the hydrogen ion concentration, hence the pH, and the relative amounts of each ionic species present in the solution.