Have you followed the rules of the Dixon's test? Have you calculated gaps and ranges?First set of volumes
25.80 27.75 25.60 25.60
25.60 25.60 25.80 27.75
gap = 27.75 - 25.80 = 1.95
range = 27.75 - 25.60 = 2.15
= 0.829 < Q = 0.906 < Q99%
= 0.926Second set of volumes
12.70 12.20 12.17 12.15
12.15 12.17 12.20 12.70
gap = 12.70 - 12.20 = 0.5
range = 12.70 - 12.15 = 0.55
= 0.829 < Q = 0.909 < Q99%
Q test is in favour of ignoring those two volumes , at 95% confidence.
Why are you so fixed up on ignoring one of the results? Any particular reason to not use them all?
Here are all the questions that has given in the assignment ...(1) What causes the lemonade to be acid? (NOTE in steps 2,3 and 4 assume that this is the only acid in the lemonade)
(2) Write a balanced equation to show the reaction of the acid in the lemonade with sodium hydroxide
(3) Using the equation of the reaction for the HCl titration and the calculation from your practical books; calculate the value exact concentration of the sodium hydroxide used. Show your calculation
(4) Using the equation of the reaction found in 2 calculate the number of moles of acid in the lemonade. How many grams of acid would be in a 500ml can of the lemonade? Show your calculations
(5) Draw a flow diagram of the exact practical process you would use to complete the above analysis.
(flow diagrams will be discussed in the Lectures)
(6) Some methods recommend that the lemonade is boiled first – explain why this might be.
(7) Could you use the same process to analyse the acid in a cola drink. If so what adjustments to the practical work would you have to make.
(8) Write a balanced equation for the titration of the acid in the cola drink with sodium hydroxide
Calculations seem to be not that much hard. So I think that they except something more than just calculations...I think that they examine whether we are good at the practical side. In practicals , errors can happen and we have to think and make choices to minimize them..I think that they have deliberately put those slightly-away volumes to make us see it and remove it from calculations.
And Also if we take that a drop of NaOH is 0.05 ml, then the guy who has done the first set of titrations has let about 20 more drops to fall and the one who has done the second set of titration has let about 10 more drops to fall accidentally.
It's just what I think about this and have no idea whether it's right.
This assignment was given at School Of Applied Sciences Bournemouth University
I'm eager to see your views..