Asked by Saira
A 2.00 ml sample of an aqueous solution of hydrogen peroxide, H2O2 (aq) is treated with an excess of Kl (aq). The liberated I2 requires 12.40 mL of 0.1025 M Na2S2O3, for its titration. Is the H2O2 up to the full strength (3% H2O2 by mass) as an antiseptic solution? Assume the density of the aqueous solution of H2O2 ( aq) is 1.00 g/ mL.
H2O2(aq) + H + I^-(aq) ---> H20(l)+I2 (not balanced)
S203^2- + I2 ----> S406 ^2- (aq) + I^- (aq) (not balanced)
H2O2(aq) + H + I^-(aq) ---> H20(l)+I2 (not balanced)
S203^2- + I2 ----> S406 ^2- (aq) + I^- (aq) (not balanced)
Answers
Answered by
GK
The balanced chemical equations are:
H<sub>2</sub>O<sub>2(aq)</sub> + 2H<sup>+</sup>(aq) + 2I<sup>-</sup>(aq) ---> 2H<sub>2</sub>0(l) + I<sub>2</sub>
2S<sub>2</sub>0<sub>3</sub><sup>2-</sup> + I<sub>2</sub> ---> S<sub>4</sub>0<sub>6</sub><sup>2-</sup>(aq) + 2I<sup>-</sup>(aq)
Moles of S2O3^2- = (0.01240L)(0.1025 mol/L) = 0.001271 S2O3^2-
Since 1 mole of I2 reacts with 2 moles of S2O3^2-,
(0.5)(0.001271mol S2O3^2-) = 0.0006355 moles I2
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1 mole of I2 is formed from 1 mole of H2O2
0.0006355 moles I2 is formed from 0.0006355 moles H2O2
(0.0006355 moles H2O2)(34g/mol) = 0.0216 g H2O2
(2.00 ml)(1.00g/mL) = 2.00g H2O2
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Now you have all the information needed to calculate the % of H2O2
H<sub>2</sub>O<sub>2(aq)</sub> + 2H<sup>+</sup>(aq) + 2I<sup>-</sup>(aq) ---> 2H<sub>2</sub>0(l) + I<sub>2</sub>
2S<sub>2</sub>0<sub>3</sub><sup>2-</sup> + I<sub>2</sub> ---> S<sub>4</sub>0<sub>6</sub><sup>2-</sup>(aq) + 2I<sup>-</sup>(aq)
Moles of S2O3^2- = (0.01240L)(0.1025 mol/L) = 0.001271 S2O3^2-
Since 1 mole of I2 reacts with 2 moles of S2O3^2-,
(0.5)(0.001271mol S2O3^2-) = 0.0006355 moles I2
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1 mole of I2 is formed from 1 mole of H2O2
0.0006355 moles I2 is formed from 0.0006355 moles H2O2
(0.0006355 moles H2O2)(34g/mol) = 0.0216 g H2O2
(2.00 ml)(1.00g/mL) = 2.00g H2O2
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Now you have all the information needed to calculate the % of H2O2