<p>If 542 microliters of 0.042 M KMnO4 are required to titrate 200.0 microliters of NaI solution to the end point in acid, what was the concentration of iodide ion in the original solution? The prodcuts are Mn (II) and I2. Write the balanced redox equations in acid.</p>
<p>K+ + MnO4- + Na+ + I- —> Mn + I2 + K+ + Na+
Get rid of spectators:
MnO4- + I- —> Mn + I2</p>
<p>Split it into half reactions:
MnO4- —> Mn
MnO4- —> Mn + 4H2O
8H+ + MnO4- —> Mn + 4H2O
Charge on left: +8 Charge on right: 0
8e- + 8H+ + MnO4- —> Mn +4 H2O</p>
<p>I- –> I2
2I- –> I2
charge on left: 2- Charge on right: 0
2I- —> I2 + 2e-</p>
<p>8e- + 8H+ + MnO4- —> Mn + 4H2O
4(2 I- —> I2 + 2 e-)</p>
<hr>
<p>8 I- + MnO4- + 8 H+ —> 4I2 + Mn + 4H2O</p>
<p>542 x 10^-6 L x (0.042 mol KMnO4 / 1 L) x (1 mol KMnO4 / 1 mol MnO4-) (1 mol MnO4- / 8 moles I-) = 2.8 x 10^-6 moles I-</p>
<p>Molarity=Moles/Liter
Molarity= (2.8 x 10^-6 moles)/ (200 x 10^-6 L) = 0.014 M I- in original solution.</p>
<p>hmm i get 0.871M, i think that your dimensionl analysis got a little messed up </p>
<p>and it should be 0.0402 not 0.042</p>
<p>You’re right! Sorry. How did I not notice that?</p>
<p>(542 x 10^-6 L KMnO4) x (0.0402 moles KMnO4 / 1 L KMnO4) x (1 mole MnO4-/1 mole KMnO4) x (8 moles I- / 1 mole MnO4-) = 1.74 x 10^-4 moles I-</p>
<p>[I-]= (1.74 x 10^-4 moles I-) / (200 x 10^-6 L) = 0.872 M I-</p>