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To a $ 25 $ ml $ { H_2O_2 } $ solution excess of acidified solution of $ { KI } $ was added. The iodine liberated require $ 20 $ ml of $ { 0.3 (N) Na_2S_2O_3 } $ solution for complete reaction.

University Name > First Year Engineering > Semester > Physical Chemistry

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$ H_2O_2 + 2KI \to 2KOH +I_2 $ $ \\ $

$ { I_2 + 2Na_2S_2O_3 \to Na_2S_4O_6 + 2NaI }\\ $

$ 20 $ ml of $ { 0.3(N) Na_2S_2O_3 = 20 × 0.3 × 10^{ –3 } } \\$ $ = 6 × 10^{ –3 } $ moles moles of $ { H_2O_2 } $ present $ = 3 × 10^{ –3 } $ mol. molarity of $ { H_2O_2 } $ solution = $ \dfrac { 3\times 10^{ -3 }}{ 25 }\times 10^3 $ $ =\dfrac 3{ 25 }( M ) $ the vol strength of $ { H_2O_2 } $ solution $ =\dfrac 3{ 25 }× 11.2 = 1.344 $ vol

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