Question 14.1: Potentiometric Precipitation Titration.
Potentiometric Precipitation Titration
A 100.0-mL solution containing 0.100 0 M NaCl was titrated with 0.100 0 M AgNO_{3}, and the voltage of the cell shown in Figure 14-7 was monitored. The equivalence volume is V_{e} = 100.0 mL. Calculate the voltage after the addition of (a) 65.0 and (b) 135.0 mL of AgNO_{3}.
The titration reaction is
Ag^{+} + Cl^{−} → AgCl(s)
(a) At 65.0 mL, 65.0% of Cl^{−} has precipitated and 35.0% remains in solution:
To find the cell voltage in Equation 14-1, we need to know [Ag^{+}]:
E = 0.558 + 0.059 16 \log[Ag^{+}] (14-1)
[Ag^{+}][Cl^{−}] = K_{sp} ⇒ [Ag^{+}] = \frac{K_{sp}}{[Cl^{−}]} = \frac{1.8 × 10^{−10}}{0.021 2 M} = 8.5 × 10^{−9} M
The cell voltage is therefore
E = 0.558 + 0.059 16 \log(8.5 × 10^{−9}) = 0.081 V
(b) At 135.0 mL, there is an excess of 35.0 mL of AgNO_{3} = 3.50 mmol Ag^{+} in a total volume of 235.0 mL. Therefore, [Ag^{+}] = (3.50 mmol)/(235.0 mL) = 0.014 9 M. The cell voltage is
E = 0.558 + 0.059 16 \log(0.014 9) = 0.450 V
Test Yourself Find the voltage after addition of 99.0 mL of AgNO_{3}. (Answer: 0.177 V)