Question 11.8: Calculating Ksp with a Concentration Cell A silver concentra...
Calculating K _{sp} with a Concentration Cell
A silver concentration cell similar to the one shown in Fig. 11.11 is set up at 25°C with 1.0 M AgNO_{3} in the left compartment and 1.0 M NaCl along with excess AgCl(s) in the right compartment. The measured cell potential is 0.58 V. Calculate the K _{sp} value for AgCl at 25°C.
In this case at 25°C
ξ = 0.58 V = 0 – \frac{0.0591}{1} \log(\frac{ [Ag ^{+}]}{1.0})
where [Ag ^{+}] represents the equilibrium concentration of Ag ^{+} in the compartment containing 1.0 M NaCl and AgCl(s). We calculate [Ag ^{+}] as follows:
\log [Ag ^{+}] = – \frac{0.58}{0.0591} = – 9.80 and [Ag ^{+}] = 1.6 × 10 ^{-10} M
Thus
K _{sp }= [Ag ^{+}] [Cl^ {-}] = (1.6 × 10 ^{-10} ) (1.0) = 1.6 × 10 ^{-10}
This calculation neglects any complications from complex ions and ion pairs.