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What is the concentration of hydrogen peroxide in run #1 when I add 10.0 ml of...

What is the concentration of hydrogen peroxide in run #1 when I add 10.0 ml of 3% of hydrogen peroxide and 35.0 ml of dI water and 15.0 ml of 0.10 M KI?

What is the concentration of hydrogen peroxide in run #2 when I add 20.0 ml of 3% of hydrogen peroxide and 25.0 ml of dI water and 15.0 ml of 0.10 M KI?

Solutions

Expert Solution

#1

We have 3 g of H2O2 in 100 g water, meaning we have 0.3 g H2O2 and 9.7 g water.

When hydrogen peroxide is added to a solution of potassium iodide, the iodide ions are slowly oxidized

according to the equation

Molecular equation: 2KI(aq) + 2HCl(aq) + H2O2 (aq) = I2 (s) + 2H2O(l)

Net ionic equation : 2I- (aq) + 2H+ (aq) + H2O2 (aq) = I2 (s) + 2H2O(l)

Molar mass of H2O2 is 34.01.

Moles of H2O2 = 0.3/34.01 = 0.0088 moles of H2O2

We have 0.1 moles of KI in 1 Ltr therefore in 15 ml we have 0.0015 moles of KI or I-

According to equation 2 moles of I- react with 1 mole of H2O2.

0.0088 moles H2O2 reacts with 0.0015*2 = 0.0030 moles of I- resulting in (0.0088-0.0030 = 0.0058 moles of H2O2 in (9.7+35+15 = 59.7) 59.7 ml water.

Molarity of H2O2 is 0.0058 moles/0.0597 L water = 0.0971 M

#2

We have 3 g of H2O2 in 100 g water, meaning we have 0.6 g H2O2 and 19.4 g water.

When hydrogen peroxide is added to a solution of potassium iodide, the iodide ions are slowly oxidized

according to the equation

Molecular equation: 2KI(aq) + 2HCl(aq) + H2O2 (aq) = I2 (s) + 2H2O(l)

Net ionic equation : 2I- (aq) + 2H+ (aq) + H2O2 (aq) = I2 (s) + 2H2O(l)

Molar mass of H2O2 is 34.01.

Moles of H2O2 = 0.6/34.01 = 0.018 moles of H2O2

We have 0.1 moles of KI in 1 Ltr therefore in 15 ml we have 0.0015 moles of KI or I-

According to equation 2 moles of I- react with 1 mole of H2O2.

0.018 moles H2O2 reacts with 0.0015*2 = 0.0030 moles of I- resulting in (0.018-0.0030 = 0.015 moles of H2O2 in (19.4+25+15 = 59.4) 59.4 ml water.

Molarity of H2O2 is 0.015 moles/0.0594 L water = 0.2525 M

queen_honey_blossom answered 2 years ago
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