Let's break this problem down **step-by-step** for each NaBr volume. We'll use the **Nernst equation** and stoichiometry for the titration. --- ## **Given:** - \( V_{\text{AgNO}_3} = 20.0 \) mL (\(0.0200\) L) - \( [\text{AgNO}_3] = 0.200 \) M - \( [\text{NaBr}] = 0.100 \) M - \( E_{\text{cell}} = 561 \) mV (at 0 mL NaBr) - Indicator: Ag/AgCl electrode ### **Step 1: Write the Reaction and Find Initial Moles** **Reaction:** \[ \text{Ag}^+ (aq) + \text{Br}^- (aq) \rightarrow \text{AgBr} (s) \] **Initial moles of AgNO₃:** \[ n_{\text{AgNO}_3} = 0.0200~L \times 0.200~mol/L = 0.00400~mol \] **For each volume of NaBr:** \[ n_{\text{NaBr}} = (\text{volume in L}) \times 0.100~mol/L \] - At 10.0 mL: \(0.0100~L \times 0.100~mol/L = 0.00100~mol\) - At 20.0 mL: \(0.0200~L \times 0.100~mol/L = 0.00200~mol\), etc. **Explanation:** We need to know how much Ag⁺ and Br⁻ we have at each stage to determine which is in excess and their concentrations. --- ### **Step 2: Find [Ag⁺] at Each Point** #### General Approach: - **Before equivalence point:** Ag⁺ in excess. - **At equivalence point:** All Ag⁺ and Br⁻ have reacted. - **After equivalence point:** Br⁻ in excess, [Ag⁺] determined by Ksp of AgBr. #### (a) **0.00 mL NaBr (initial):** - Only Ag⁺ present: \([Ag^+] = 0.200~M\). - \(E_{\text{cell}} = 561~mV\) (already given). #### (b) **10.0 mL NaBr:** - Moles NaBr: \(0.00100~mol\) - Moles Ag⁺ left: \(0.00400 - 0.00100 = 0.00300~mol\) - Total volume: \(20.0 + 10.0 = 30.0~mL = 0.0300~L\) - \([Ag^+] = 0.00300~mol / 0.0300~L = 0.100~M\) #### (c) **20.0 mL NaBr:** - Moles NaBr: \(0.00200~mol\) - Moles Ag⁺ left: \(0.00400 - 0.00200 = 0.00200~mol\) - Total volume: \(40.0~mL = 0.0400~L\) - \([Ag^+] = 0.00200~mol / 0.0400~L = 0.0500~M\) #### (d) **40.0 mL NaBr:** - Moles NaBr: \(0.00400~mol\) - Equivalence point reached (\(0.00400~mol\) Ag⁺, \(0.00400~mol\) Br⁻). - After eq. point, Br⁻ in excess. - \(n_{\text{Br}^-} = 0.00400~mol\) (from NaBr) - \(n_{\text{Ag}^+} = 0\) (all reacted) After equivalence, [Ag⁺] is determined by Ksp: \[ \text{AgBr} (s) \leftrightarrow \text{Ag}^+ + \text{Br}^- \] \[ K_{sp} = [\text{Ag}^+][\text{Br}^-] \] \(K_{sp} (\text{AgBr}) \approx 5.0 \times 10^{-13}\) - Moles excess Br⁻: \(0.00400~mol\) (from NaBr) - Total volume: \(60.0~mL = 0.0600~L\) - \([\text{Br}^-] = 0.00400~mol / 0.0600~L = 0.0667~M\) \[ [\text{Ag}^+] = \frac{K_{sp}}{[\text{Br}^-]} = \frac{5.0 \times 10^{-13}}{0.0667} \approx 7.5 \times 10^{-12}~M \] #### (e) **50.0 mL NaBr:** - Moles NaBr: \(0.00500~mol\) - Excess Br⁻: \(0.00500 - 0.00400 = 0.00100~mol\) - Total volume: \(70.0~mL = 0.0700~L\) - \([\text{Br}^-] = 0.00100~mol / 0.0700~L \approx 0.0143~M\) \[ [\text{Ag}^+] = \frac{5.0 \times 10^{-13}}{0.0143} \approx 3.5 \times 10^{-11}~M \] **Explanation:** We use stoichiometry before the equivalence point and \(K_{sp}\) after equivalence to calculate [Ag⁺]. --- ### **Step 3: Use the Nernst Equation to Calculate \(E_{\text{cell}}\)** **Half-cell reaction:** \[ \text{AgCl} (s) + e^- \rightarrow \text{Ag} (s) + \text{Cl}^- \] But in this setup, the indicator electrode is Ag, and the reference is Ag/AgCl. **Simplified Nernst equation for Ag⁺/Ag:** \[ E = E^\circ_{\text{Ag}^+/\text{Ag}} - 0.0592 \log [\text{Ag}^+] \] But the reference electrode (Ag/AgCl) potential is constant, so \(E_{\text{cell}}\) changes only with [Ag⁺]. Let’s use the initial data to calibrate: \[ 561~mV = E^\circ_{\text{cell}} - 0.0592 \log(0.200) \] Let \(E^\circ_{\text{cell}}\) be the combined standard potential: \[ E^\circ_{\text{cell}} = 561~mV + 0.0592 \log(0.200) \] \[ \log(0.200) = -0.6990 \] \[ 0.0592 \times (-0.6990) \approx -0.0413 \] \[ E^\circ_{\text{cell}} = 561 + 41.3 = 602~mV \] Now, calculate \(E_{\text{cell}}\) at each step: - For each, \(E_{\text{cell}} = 602~mV - 0.0592 \log([\text{Ag}^+])\) --- ## **Final Answers (Plug in the values):** ### **0.00 mL NaBr:** \[ E_{\text{cell}} = 602 - 0.0592 \log(0.200) = 561~mV \] (Already provided, matches our calibration.) ### **10.0 mL NaBr:** \[ E_{\text{cell}} = 602 - 0.0592 \log(0.100) \] \[ \log(0.100) = -1.000 \] \[ E_{\text{cell}} = 602 - [0.0592 \times (-1)] = 602 + 0.0592 = 661~mV \] ### **20.0 mL NaBr:** \[ E_{\text{cell}} = 602 - 0.0592 \log(0.0500) \] \[ \log(0.0500) = -1.301 \] \[ E_{\text{cell}} = 602 - [0.0592 \times (-1.301)] = 602 + 0.0770 = 679~mV \] ### **40.0 mL NaBr:** \[ E_{\text{cell}} = 602 - 0.0592 \log(7.5 \times 10^{-12}) \] \[ \log(7.5 \times 10^{-12}) = \log(7.5) + \log(10^{-12}) = 0.8751 - 12 = -11.125 \] \[ E_{\text{cell}} = 602 - 0.0592 \times (-11.125) = 602 + 0.6585 = 660.8~mV \approx 661~mV \] ### **50.0 mL NaBr:** \[ E_{\text{cell}} = 602 - 0.0592 \log(3.5 \times 10^{-11}) \] \[ \log(3.5 \times 10^{-11}) = 0.5441 - 11 = -10.456 \] \[ E_{\text{cell}} = 602 - 0.0592 \times (-10.456) = 602 + 0.6189 = 620.9~mV \approx 621~mV \] --- ## **Summary Table:** | NaBr (mL) | [Ag⁺] (M) | Ecell (mV) | |-----------|-------------------|------------| | 0.00 | 0.200 | 561 | | 10.0 | 0.100 | 661 | | 20.0 | 0.0500 | 679 | | 40.0 | 7.5 × 10⁻¹² | 661 | | 50.0 | 3.5 × 10⁻¹¹ | 621 | --- ### **Final Answer Table:** | NaBr (mL) | Ecell (mV) | |-----------|------------| | 0.00 | 561 | | 10.0 | 661 | | 20.0 | 679 | | 40.0 | 661 | | 50.0 | 621 | --- **Each step involved stoichiometry, equilibrium, and the Nernst equation to find the cell voltage at each point.**

The problem combines concepts of titration, electrochemical cell potentials, and the Nernst equation. First, it involves stoichiometric calculations to determine the amount of silver ion remaining at different stages of titration based on the volume of added NaBr, considering the initial concentration and volume of AgNO₃. Next, it applies the solubility product (Ksp) to find the concentration of Ag⁺ after the equivalence point when excess Br⁻ is present. Finally, the Nernst equation is used to relate the measured cell potential to the concentration of Ag⁺ ions, allowing calculation of the cell voltage at each titration point. These combined concepts enable tracking the cell potential throughout the titration process.