Answer the entirety of the question: a, b, c, d, e, and f. (This is a pre-lab) H
ID: 508703 • Letter: A
Question
Answer the entirety of the question: a, b, c, d, e, and f. (This is a pre-lab)
How do you expect increasing the salt concentration to affect the final concentration of bleach? b. How do you expect increasing the amount of current to affect the final concentration of bleach? c. Calculate the theoretical yield(in molarity) if the electrolysis reaction was run using 80mL of 5M NaCl solution and 0.600A for 30minutes d. A total of 8.00mL of solution was used to titrate a bleach solution to the endpoint. What are the molarity, ppm, and % by mass of the bleach? e. What was the percent yield of the above reaction? f What mass of NaCl solid should be added to 100mL of pure water to produce a 6.00M solution?Explanation / Answer
a)
Increasing the salt concentration increases electron mobility, which in turn increases rate of reaction.
It will also increase the final concentration of bleach.
b)
As per Faraday’s electrolysis law, more electrons = more product.
Increased current will definitely increase rate of reaction. But the number of molecules is same.
Increased current will not affect final concentration of bleach.
c)
First we will need to calculate number of electrons
No. of electrons = n = It/F
= 0.600A*1800sec / 96485 (Faraday’s constant) = 0.011 electrons
Now, electrolysis reaction is,
ClO-(aq) + Na+ = NaClO (aq) means moles of Na+ determines the yield.
58.456 (m.w of NaCl) * 5 moles * 0.1 litre = 29.29 g in water to a final volume of 100 ml
But we have 80 mL water. Therefore, 29.29*0.8 = 23.432 g NaCl
23.432 g NaCl * 1 mole/58.456 g * 1 mole/1 mole * 74.44 g NaClO = 29.83 g
Theoretical yield is 29.83 g
d)
[I3-] [starch]+ 2S2O3 2- = 3I- + S4O6 2- + starch
Three moles of iodine react with 2 moles of thiosulphate. Also, same amount react with 1 mole of bleach
1 mole bleach * 3 moles iodine = 2 moles thiosulfate * 3 moles iodine
Cancelling iodine from both sides
1 mole bleach = 2 moles thiosulphate
[bleach] = ½ * [thiosulphate] * Vol(thiosulphate) / V(bleach)
V(bleach) = ½ * [thiosulphate] * Vol(thiosulphate) / [bleach]
We will consider both concentration as 0.1 mol/L
= ½ *0.1*8/0.1 = 4 mL
Molarity is assumed as 0.1 mol/L. = 7.44 g/L = 7440 mg/L = 7440 ppm