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Lera25 [3.4K]
4 years ago
10

When standardizing an iodine solution, 30.37 mL of the solution was required to

Chemistry
1 answer:
dexar [7]4 years ago
6 0

<span>First thing, we need to know the chemical equation for the titration reaction and it is as follows:

KIO3(aq) + 6 H+ (aq) + 5 I- (aq) = 3 I2(aq) + 3 H2O(l) + K+ (aq) </span>
<span> C6H8O6(aq) + I2(aq) = C6H6O6(aq) + 2 I- (aq) + 2 H+ (aq)</span>

We are given the amounts and the concentrations. We use these values to calculate what is needed. We calculate as follows:

 

<span>5 mg/mL C6H8O6 ( 10.0 mL ) ( 1 g / 1000 mg ) (1 mol / 176.14 g) ( 1 mol I2 / 1 mol C6H8O6 ) ( 1 mol KIO3 / 3 mol I2 ) = 9.46x10^-5 mol KIO3 needed</span>

 

<span>Molarity = 9.46x10^-5 mol KIO3 / .03037 L = 0.003 M KIO3</span>

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