Question AE.2: Using Normality How many grams of potassium oxalate should b...
Using Normality
How many grams of potassium oxalate should be dissolved in 500.0 mL to make a 0.100 N solution for titration of MnO_4^{−} ?
5H_{2}C_{2}O_{4} + 2MnO_{4}^{−} + 6H^{+} \rightleftharpoons 2Mn^{2+} + 10CO_{2} + 8H_{2}O (E-5)
It is first necessary to write the oxalic acid half-reaction:
H_{2}C_{2}O_{4} \rightleftharpoons 2CO_{2} + 2H^{+} + 2e^{−}
It is apparent that there are two equivalents per mole of oxalic acid.
Hence, a 0.100 N solution will be 0.050 0 M:
\frac{0.100 \cancel{equiv}/L}{2 \cancel{equiv}/mol} = 0.050 0 mol/L = 0.050 0 M
Therefore, we must dissolve (0.050 0 mol/\cancel{L})(0.500 0 \cancel{L}) = 0.025 0 mol in 500.0 mL. Because the formula mass of K_{2}C_{2}O_{4} is 166.216, we should use (0.025 0 mol) × (166.216 g/mol) = 4.15 g of potassium oxalate.