Posted by Gloria on Tuesday, July 30, 2013 at 8:55am.
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 - Graham, Tuesday, July 30, 2013 at 9:26am
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 - Gloria, Tuesday, July 30, 2013 at 9:47am
Thank you so so so much! You're great!
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