dropper of a 3% starch solution. blue-black color, indicating the presence of th
ID: 712966 • Letter: D
Question
dropper of a 3% starch solution. blue-black color, indicating the presence of the starch-la complex care, dropwise, until the solution turns colorless. At this point the The solution will change to a Continue the titration with titration is complete (all the I2 has reacted). Record the final buret reading. Subtract the two buret readings to obtain the total volume of sodium thiosulfate solution used for the titration. Repeat the titration on fresh bleach samples of the same brand at least two more times. You must do at least 3 titrations. Report 1. Calculate the molarity of your standard sodium thiosulfate pentahydrate solution. 2. For each of the three bleach samples: How many moles of sodium thiosulfate were used in the titration? a. b. How many moles of l2 must have reacted with this amount of sodium thiosulfate? c. How many moles of NaOCI must have produced the amount of 12 calculated in (b)? d. How many grams of NaOCI were in the sample of bleach? e. What is the percentage by mass of NaOCI in the bleach sample? 3. Calculate the average percent NaOCl in your brand of bleach. 4. Calculate the average deviation and the RAD and comment on your precision. 5. Is your average percent NaOCI close to the percentage NaOCI advertised on commercial brands? List a few. he hle4 f.sk:8.wa at vl: 37.sb empty Fusk: 127143 mass of beach: 2.268 1·632 1.779 31.Adnl 8.30 m weghted flak 129.0 Finl vl: 34.10 Trixl 3 enpy 52Explanation / Answer
1) Molarity = (number of moles) / (volume in litres) & number of moles =(given mass) / (molar mass) = 6.241 / 248
Molarity = (6.241 / 248) / volume in L
(you did not specify the volume of Na2S2O3.5H2O solution prepared, therefore I cannot unfortunately evaluate the exact values for the entire question. still, I tried my best to put it in a way for you to understand the concept and calculations easily)
2) For each trial
a) moles of Na2S2O3.5H2O used = molarity X volume used in that trial = (molarity as calculated in part 1) X V
b) gram equivalents of sodium thiosulfate used = gram equivalents of I2 used
which is equal to M X V X n-factor ( all of sodium thiosulfate ) = M X V X 1 ( 1 is the n-factor)
and we know that number of moles = number of gm equivalents / n-factor of I2 = (M X V X 1) / 2
c) the reaction of NaOCl with KI to produce I2 is given as :
I2(aq)
It can be seen that 1 gm equivalent of I2 is produced by 1 gm equivalent of ClO-
hence, no of moles of ClO- used = number of equivalents of I2 produced as calculated above / n-factor of NaOCl which is 1
d) grams of NaOCl in the sample = (number of moles of naOCl as calculated above) X (molar mass of NaOCl (74.4))
e) percentage by mass of NaOCl in bleach sample = (mass of NaOCl (in grams) calculated above) / mass of bleach
3) average mass of bleach = average of the three recorded values = (2.268 + 1.632 + 1.974) / 3 = 1.958 g
average mass of NaOCl = average of the three calculated values for part d
average percent mass = average mass of NaOCl / average mass of bleach
4) a commercial bleach brand generally contains about 3.62 mass % of NaOCl. use this value to compare with your calculated value.
I2(aq)
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