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(c) Potassium hydroxide reacts completely with sulfuric acid according to the equation:
H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l)
0.95 g of potassium hydroxide were dissolved in water and made up to 250 cm3 of solution. 25 cm3 samples of this solution were titrated against sulfuric acid (H2SO4) solution. The mean volume (titre) of sulfuric acid required for complete reaction was 15.00 cm3.
(i) Find the molar concentration of the prepared solution of potassium hydroxide. Complete the calculation given below:
Relative Molecular Mass (RMM) of KOH =

= ………………… (1 mark)
No. of moles of KOH in 250 cm3 =
= ………………………… (1 mark)
molar concentration of KOH =
(Include units in your answer) = ………………………… (1 mark)



Continued…..

(ii) Find the molar concentration of the sulfuric acid from the titration result:
Complete the calculation given below:
No. of moles of KOH in 25.00 cm3 =
= ………………………… (1 mark)
no. of moles H2SO4 in 15.00 cm3 =
= ………………………… (1 mark)
molar concentration of H2SO4 =
(Include units in your answer) = ………………………… (2 marks)

(Total 10 marks)

Continued…..

Section B: Biological Science - Two questions (max 30 marks).
Make sure you have read materials on Canvas FX3003 within T8.
Q4.
A student conducts an enzyme experiment revolving around changing the enzyme concentration, and its effects upon the rate of reaction when applied to the enzyme amylase and starch in the presence of a pH Buffer of 7. The student obtains a data set based upon the time taken for the disappearance of starch, associated with a sustained colouration of brown when the mixture was added to a spotting tile containing I/KI (iodine in potassium iodide), having mixed equal volumes of standard starch with a given concentration of enzyme.
The student obtains the following results: it took 1475 seconds for the disappearance of starch when the 0.05% amylase was used, 12 minutes and 15 seconds for the 0.1% amylase, whilst the 0.15% amylase took 520 seconds. The 0.25% amylase gave a result of four minutes and 35 seconds, whilst the 0.2% amylase took 6 minutes and five seconds.
The student is informed that as the results reflect the use of the disappearance of substrate and not the production of product, their results must be converted to 1/T (where T=time in seconds) for the rate of reaction.

a) Help the student by processing the data and producing a suitable table of results.
(6 marks)

b) From the results produce a suitable graph - in excel (or other suitable package) and insert the graph here. (You may draw by hand the graph on suitable graph paper and scan and insert as an alternative to the use of excel). (6 marks)

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  1. (c) H2SO4(aq) + 2KOH(aq) ==> K2SO4(aq) + 2H2O(l)
    (i) RMM KOH = 39.1 + 1 + 16 = 56.1
    mols KOH = grams/molar mass = 0.95/56.1 = 0.0169 mols KOH in 250 cc.
    (KOH) = 0.0169/0.250 L = 0.0677 mols KOH/L solution.
    (ii) mols KOH in 25 cc is
    0.0677 x (25 cc/1000 cc) = 0.00169 mols KOH
    mols H2SO4 in 15 cc = mols KOH x (1 mol H2SO4/2 mols KOH) = 0.00169 x 1/2 = 0.000846.
    (H2SO4) = mols H2SO4/L H2SO4 = 0.000846 mols/0.015 cc = 0.0564 mols/L = 0.0564 M
    B.
    a. Surely you can process the data on your own. It's just a conversion.
    b. Sorry, but this site does not support drawing structures, diagrams, graphs, etc.

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    DrBob222

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