1989 WAEC Chemistry Practical D is 2.00M ethanoic acid.  E is 2.00M potassium hydroxide solution. (a) Using a 50cm\(^3\) measuring cylinder,...

WAEC 1989

D is 2.00M ethanoic acid.  E is 2.00M potassium hydroxide solution.

(a) Using a 50cm\(^3\) measuring cylinder, measure 50cm\(^3\) of D and transfer the solution into plastic cup.Record the temperature T\(_1\) of the solution. Rinse the cylinder distilled water and allow to dry.

 

(b) Using the dry measuring cylinder from (a) above, measure 50cm of E. Record the temperature, T\(_2\) of the solution.

 

(c) Find the average temperature T\(_3\) of the two solutions and record the value.

 

(d) Pour the measured quantity of solution E quickly from the measuring cylinder into the plastic cup containing solution D. Stir the mixture with the thermometer. Record the highest temperature T\(_4\) attained.

(e)(i) Find the rise in temperature (T\(_4\) - T\(_3\) ad record the value 

(ii) Calculate the mass of the reaction mixture, given that during the reaction 1cm\(^3\) of the mixture weighs 1g

 

(f) From your results in (a) to (e) above, calculate the;

 (i) Heat evolved during the reaction, giving that the specific heat capacity of water is 4.2Jg\(^{-1}\)C \(^{-1}\) and using the formula; heat evolved = mass x specific heat capacity x rise in temperature

(ii) Heat of neutralization of one mole of ethanoic acid by potassium hydroxide

(g) List two sources of error in the method used for determining the heat of neutrailzation and suggest how their effect can be minimized. 

 

Explanation

Solution D is 2M CH\(_3\)COOH. and solution E of solution E is 2M KOH 

(a) Temp. T\(_1\) of solution D = 31\(^o\)C 

(b) Temp. T\(_2\) of solution E = 32\(^o\) 

(c) Average Temperature T\(_3\) = \(\frac{T_1 + T_2}{2} = \frac{31 + 32}{2}\)  = 31.50\(^o\)C

(d) The highest temp. T\(_4\) = 40\(^o\) 

(e)(i) Rise in temp. T\(_4\) - T\(_3\) = 40 - 31.50 = 8.50\(^o\)C 

(ii) 50cm\(^3\) of soln D + 50cm\(^3\) of soln. E = 100cm\(^3\) of mixture.1cm\(^3\) of solution weighs 1g. (given) 

100cm\(^3\) of the mixture will weigh 1 x 100 = 100g 

 

(f)(i) heat evolved = mass x specific heat capacity x rise in temperature = 100 x 4.2 x 8.5 joules 

= 3570 joules 

(ii) nos. of moles = \(\frac{volume \times mol}{100}\) 

for the acid; \(\frac{50 \times 2}{1000}\) = 0.10 moles 

for the base = \(\frac{50 \times 2}{1000}\) = 0.10 moles 

Equation for the reaction; CH\(_3\)COOH\(_{(aq)}\) + KOH\(_{(aq)}\) \(\to\) H\(_2\)O\(_{(l)}\)

From the aqueous neutralisation above, 1 mole of CH\(_3\)COOH + 1 mole of KOH produces one mole of water 

0.1 mole of CH\(_3\)COOH will produce 0.1 mole water. Hence 0.10 mole of H\(_2\)O is produced with the evolution of 3570 joules of heat energy 

1 mole of H\(_2\)O will be produced with the evolution of \(\frac{350}{0.1}\) J mol\(^{-1}\) = 35700 Jmol\(^{-1}\) = 35.70KJ mol\(^{-1}\)

The heat of neutralisation 35.70KJ water formed,

(g) Sources of errors.

(i) Heat is lost 

(ii) Density of solutions not exactly 1 gcm\(^{-3}\) 

(iii) Thermometer may not be accurately calibrated 

(iv) Loss of soln. during transfer and stirring.

 

To minimize error;

(i) Use lagged  calorimeter 

(ii) Find water equivalent of calorimeter 

(iii) Measure mass of solution and mixture accurately.

(iv) Careful string and transfer 

 


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