Question 14.E.A: The apparatus in Figure 14-7 was used to monitor the titrati...
The apparatus in Figure 14-7 was used to monitor the titration of 50.0 mL of 0.100 M AgNO_{3} with 0.200 M NaBr. Calculate the cell voltage at each volume of NaBr, and sketch the titration curve: 1.0, 12.5, 24.0, 24.9, 25.1, 26.0, and 35.0 mL.
The reaction at the silver electrode (written as a reduction) is
Ag^{+} + e^{−} \rightleftharpoons Ag(s), and the cell voltage is written as
E = E_{+} − E_{−} = E_{+} − E(S.C.E) = E_{+} − 0.241
= (0.799 − 0.059 16 \log \frac{1}{[Ag^{+}]}) − 0.241
= 0.558 + 0.059 16 \log [Ag^{+}]
Titration reaction: Br^{−} + Ag^{+} → AgBr(s) K_{sp} = 5.0 × 10^{−13}
The equivalence point is V_{e} = 25.0. Between 0 and 25 mL, there is unreacted Ag^{+} in the solution.
1.0 mL: [Ag^{+}] = \underset{\begin{matrix} \nearrow \\ Fraction of Ag^{+} \\ remaining \end{matrix} }{(\frac{24.0}{25.0})} \underbrace{(0.100 M)}_{\begin{matrix} Initial concentration \\ of Ag^{+} \end{matrix} } \underset{\begin{matrix} \nwarrow \\ Dilution \\ factor \end{matrix} }{(\frac{50.0}{51.0})} = 0.094 1 M
⇒ E = 0.558 + 0.059 16 \log[0.094 1] = 0.497 V
12.5 mL: [Ag^{+}] = (\frac{12.5}{25.0})(0.100 M)(\frac{50.0}{62.5}) = 0.040 0 M
⇒ E = 0.475 V
24.0 mL: [Ag^{+}] = (\frac{1.0}{25.0})(0.100 M)(\frac{50.0}{74.0}) = 0.002 70 M
⇒ E = 0.406 V
24.9 mL: [Ag^{+}] = (\frac{0.10}{25.0})(0.100 M)(\frac{50.0}{74.9}) = 2.67 × 10^{−4} M
⇒ E = 0.347 V
Beyond 25 mL, all AgBr has precipitated and there is excess Br^{−} in solution.
25.1 mL: [Br^{−}] = (\frac{0.1}{75.1})(0.200 M) = 2._{67} × 10^{−4} M
⇒ [Ag^{+}] = K_{sp}/[Br^{−}] = (5.0 × 10^{−13})/(2._{67} × 10^{−4}) = 1._{88} × 10^{−9} M
⇒ E = +0.042 V
26.0 mL: [Br^{−}] = (\frac{1.0}{76.0})(0.200 M) = 2.6_{6} × 10^{−4} M
⇒ [Ag^{+}] = 1.9_{0} × 10^{−10} M ⇒ E = −0.017 V
At 35.0 mL: [Br^{−}] = (\frac{10.0}{85.0})(0.200 M) = 0.023 5 M
⇒ [Ag^{+}] = 2.1_{2} × 10^{−11} M ⇒ E = −0.073 V
