Question 5.10: Standardizing a Solution for Use in Redox Titrations. A piec......

Analyze

The key to a titration calculation is that the amounts of two reactants consumed in the titration are stoichiometrically equivalent—neither reactant is in excess. We are given a mass of Fe (0.1568 g) and must determine the number of moles of KMnO_{4} in the 26.24 mL sample. The following conversions are required:

g Fe → mol Fe → mol Fe^{2+} → mol  MnO_{4}^{−} → mol  KMnO_{4}

The third conversion, from mol Fe^{2+}  to  mol  MnO_{4}^{−}, requires a stoichiometric factor constructed from the coefficients in equation (5.29).

Solve

First, determine the amount (in moles) of KMnO_{4} consumed in the titration.

? mol KMnO_{4} = 0.1568  g  Fe × \frac{1  mol  Fe}{55.847  g  Fe} × \frac{1  mol  Fe^{2+}}{1  mol  Fe} × \frac{1  mol  MnO_{4}^{−}}{5  mol  Fe^{2+}} × \frac{1  mol  KMnO_{4}}{1  mol  MnO_{4}^{−}}

= 5.615 × 10^{−4}  mol  KMnO_{4}

The volume of solution containing the 5.615 × 10^{−4}  mol   KMnO_{4} is 26.24 mL =0.02624 L, which means that

concn KMnO_{4} = \frac{5.615  ×  10^{−4}  mol  KMnO_{4}}{0.02624   L} = 0.02140  M  KMnO_{4}

Assess

For practical applications, such as for titrations, we use solutions with molarities that are neither very large nor very small. Typically, the molarities lie in the range 0.001 M to 0.1 M. If you calculate a molarity that is significantly larger than 0.1 M, or significantly smaller than 0.001 M, then you must carefully check your calculation for possible errors.