Let's solve the problem step by step as requested: --- ### **Step 1: Calculating [OH⁻] added to the buffer solution** The equation to calculate the concentration of [OH⁻] ions added is: \[ [\text{OH}^-] = \frac{n \text{ (mol)}}{V \text{ (volume)}} \] Given: - \( n = 0.01 \) mol - \( V = 0.5 \) L (since 500 mL = 0.5 L) Plug these values into the equation: \[ [\text{OH}^-] = \frac{0.01}{0.5} = 0.02 \text{ M} \] **Explanation:** We use the formula for concentration, which is moles of solute divided by volume of solution in liters. Here, 0.01 mol of OH⁻ is added to 0.5 L of solution, giving a concentration of 0.02 M. --- ### **Step 2: Completing the Reaction Table** **Initial concentrations:** - RCOOH: 0.35 M - OH⁻: 0.02 M (from above) - RCOO⁻: 0.65 M **Reaction:** RCOOH + OH⁻ → RCOO⁻ + H₂O **Changes:** - OH⁻ will react completely with RCOOH (since both are present in the same amount, 0.02 mol in 0.5 L): - RCOOH: -0.02 M (decreases) - OH⁻: -0.02 M (decreases) - RCOO⁻: +0.02 M (increases) **Final concentrations:** - RCOOH: 0.35 - 0.02 = 0.33 M - RCOO⁻: 0.65 + 0.02 = 0.67 M **Explanation:** OH⁻ reacts with RCOOH in a 1:1 ratio, so both decrease by the same amount (0.02 M), while RCOO⁻ increases by the same amount. --- ### **Step 3: Calculating the pH of the solution** This is a buffer solution, so we use the Henderson-Hasselbalch equation: \[ \text{pH} = \text{p}K_a + \log \left(\frac{[\text{RCOO}^-]}{[\text{RCOOH}]}\right) \] Given: - \( K_a = 5.8 \times 10^{-5} \implies \text{p}K_a = -\log(5.8 \times 10^{-5}) \approx 4.24 \) - \([\text{RCOO}^-] = 0.67\) M - \([\text{RCOOH}] = 0.33\) M Calculate pH: \[ \text{pH} = 4.24 + \log\left(\frac{0.67}{0.33}\right) = 4.24 + \log(2.03) \approx 4.24 + 0.31 = 4.55 \] **Explanation:** The Henderson-Hasselbalch equation is used for buffer solutions; we insert the final concentrations and calculated pKa to find the pH. --- ## **Final Answer:** ### pH = **4.55**

**Step 1: Calculate [OH⁻] added to the solution** - Moles of solid LiOH added = 0.01 mol - Volume of solution = 500 mL = 0.5 L \[ [\text{OH}^-] = \frac{\text{moles of OH}^-}{\text{volume in liters}} = \frac{0.01}{0.5} = 0.02\, \text{M} \] *Explanation:* The concentration of hydroxide ions introduced is directly obtained by dividing the moles added by the total solution volume. --- **Step 2: Determine the new concentrations of the buffer components after reaction** Initial concentrations: - RCOOH = 0.35 M - RCOONa (RCOO⁻) = 0.65 M - [OH⁻] added = 0.02 M Reaction: \[ \text{RCOOH} + \text{OH}^- \rightarrow \text{RCOO}^- + \text{H}_2\text{O} \] - OH⁻ reacts completely with RCOOH in a 1:1 molar ratio: - RCOOH decreases by 0.02 M: \(0.35 - 0.02 = 0.33\, \text{M}\) - RCOO⁻ increases by 0.02 M: \(0.65 + 0.02 = 0.67\, \text{M}\) *Explanation:* The hydroxide ions neutralize an equivalent amount of RCOOH, converting it into RCOO⁻, thus altering their concentrations while water is formed. --- **Step 3: Calculate the pH using the Henderson-Hasselbalch equation** - \(K_a = 5.8 \times 10^{-5}\) - \(\text{p}K_a = -\log(5.8 \times 10^{-5}) \approx 4.24\) Insert values: \[ \text{pH} = 4.24 + \log \left(\frac{0.67}{0.33}\right) = 4.24 + \log(2.03) \approx 4.24 + 0.31 = 4.55 \] *Explanation:* Using the Henderson-Hasselbalch equation with the updated buffer concentrations gives the final pH of the solution. --- ### **Final Answer: The pH of the solution is approximately 4.55**

1. **Calculate [OH⁻] added:** \[ \text{[OH}^-] = \frac{0.01\, \text{mol}}{0.5\, \text{L}} = 0.02\, \text{M} \] *The hydroxide concentration after addition is 0.02 M.* 2. **Determine new buffer concentrations:** - RCOOH: \(0.35 - 0.02 = 0.33\, \text{M}\) - RCOO⁻: \(0.65 + 0.02 = 0.67\, \text{M}\) *The buffer components adjust due to the neutralization reaction.* 3. **Calculate pH using Henderson-Hasselbalch:** \[ \text{pH} = 4.24 + \log \left(\frac{0.67}{0.33}\right) \approx 4.24 + 0.31 = \boxed{4.55} \] *Final pH of the solution is approximately 4.55.*

Certainly! Here's a concise, step-by-step summary with the final answers: **a) Equation used to calculate [OH⁻]:** \[ [\text{OH}^-] = \frac{\text{moles of OH}^-}{\text{volume in liters}} \] *(Choice: concentration = moles/volume)* **b) Final concentrations after reaction:** - RCOOH: \(0.35\, \text{M} - 0.02\, \text{M} = 0.33\, \text{M}\) - RCOO⁻: \(0.65\, \text{M} + 0.02\, \text{M} = 0.67\, \text{M}\) **c) Final pH of the solution:** \[ \boxed{\text{pH} \approx 4.55} \]

The problem involves calculating the pH of a buffer solution after adding a strong base. First, we determine the concentration of hydroxide ions introduced by the solid LiOH using the molarity formula (moles divided by volume). Next, we account for the reaction between OH⁻ and the weak acid RCOOH, which neutralizes part of the acid and converts it into its conjugate base RCOO⁻, updating their concentrations accordingly. Finally, we use the Henderson-Hasselbalch equation, which relates pH to the ratio of conjugate base and weak acid concentrations and the acid's pKa, to find the final pH of the buffer after the reaction.

The problem involves understanding buffer chemistry and the effect of adding a strong base to a buffer solution. The first step is to calculate the concentration of hydroxide ions ([OH⁻]) introduced by the solid LiOH, which is done by dividing the number of moles added by the total volume of the solution, using the molarity formula. This provides the initial amount of base available to react with the weak acid in the buffer. Next, the reaction between the added hydroxide ions and the weak acid (RCOOH) is considered; since they react in a 1:1 molar ratio, the concentrations of RCOOH decrease, and the conjugate base RCOO⁻ increases by the same amount. These new concentrations are calculated by subtracting and adding the reacted molarity from the initial concentrations. Finally, the pH of the resulting buffer solution is determined using the Henderson-Hasselbalch equation, which relates the pH to the pKa of the weak acid and the ratio of the concentrations of its conjugate base to the weak acid. This approach allows us to find the final pH after the neutralization reaction, considering the buffer capacity and the added base.

**a)** The equation used to calculate [OH⁻] is: \[ [\text{OH}^-] = \frac{\text{moles of OH}^-}{\text{volume in liters}} \] **b)** Final concentrations after reaction: - RCOOH: **0.33 M** - RCOO⁻: **0.67 M** **c)** Final pH: **approximately 4.55**