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March 29, 2017

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A 25 ml sample of household bleach was diluted to 500 ml in a volumetric flask. An unmeasured excess of potassium iodide was aded to a 20 ml aliquot of the diluted sample; the iodine liberated in the reaction: OCl- + 2I- -> I2 + Cl- + H2O
required 34.5 ml of 0.0409 M Na2S2O3. Calculate the weight- volume precentage of NaOCl in the sample.

i just get stuck in the middle of the problem :S

  • Analytical chemistry - ,

    The titration is: 2 Na2S2O3 + I2 = Na2S4O6 + 2 NaI

    moles NaOCl in 20mL aloquat
    = moles I2
    = (1/2)moles Na2S2O3
    = 0.345 L * 0.0409 M / 2

    moles NaOCl in original (25mL) sample
    = 0.0345 * 0.0409 * (500/20) mol / 2

    mass NaOCl in original (25mL) sample
    = 74.4424 g/mol * 0.0345 L * 0.0409 M * 12.5

    %w/v of NaOCl in original sample
    = gram nass solute per 100mL solution
    = 74.4424 * 0.0345 * 0.0409 * 12.5 * 4 g/100mL
    = 5.25... g/100mL
    = 5.25... %w/v

  • Analytical chemistry - ,

    Thank you so so so much! You're great!

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