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2003 WAEC Chemistry Practical All your burette readings (initial and final), as well as the size of your pipette, must...

Chemistry
WAEC 2003

All your burette readings (initial and final), as well as the size of your pipette, must be recorded but no account of experimental procedure is required. All calculations must be done in your answer book. A is a solution of HCl containing 7.30g dm\(^{-3}\), B is a solution of X\(_2\)CO\(_{3}\) containing 10.6 gdm\(^{-3}\)

(i) Put A into your burette and titrate readings against 20.0 cm\(^3\) or 25.0cm\(^3\) portions of B using methyl orange as indicator. Tabulate your burette reading and calculate the average volume of A used. The equation for the reaction involved in the titration is ;

X\(_2\)CO\(_{3(aq)}\) + 2HCl\(_{(aq)}\) \(\to\) 2XCl\(_{(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)

(ii) From your results and the information provided above, calculate the (i) concentration or A in mol dm\(^{3-}\)

 

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Explanation

Volume of pipette used = 25.00 cm\(^3\). Indicator used; Methyl Orange 

Colour change at the endpoint from orange for pink/red

Burette readings Rough 1st acc. 2nd acc. 3rd acc.
Final reading in cm\(^3\) 23.00 46.50 23.40 46.80
Initial reading in cm\(^{3}\)  0.00 23.00 0.00 23.40
Volume of A (acid) used 23.00 23.50 23.40 23.40

 

Average volume of acid used = \(\frac{23.50 + 23.40 + 23.40}{3}\) = 23.43 cm\(^3\)

 

(b) Equation of the reaction is; X\(_2\)CO\(_{3(aq)}\) + 2HCl\(_{3(aq)}\) \(\to\) 2Cl\(_{(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)

(i) To calculate the concentration of A. Molar mass of HCl = 36.5 gmol\(^{-1}\) 

Concentration of A (C\(_A\)) = \(\frac{7.30}{36.5}\) = 0.200 mol dm\(^{-3}\)

(ii) Tp calculate the concentration of solution B; from the equation, mole ratio of acid is 2;1 

\(\frac{C_AV_A}{C_BV_B} = \frac{2}{1}\)

= \(\frac{0.2 \times 23.43}{C_B \times 25} = \frac{2}{1}\)

C\(_B\) = \(\frac{0.2 \times 23.43}{2 \times 25}\)

= 0.093

Approx. 0.100 mol dm\(^3\) 

(iii) To calculate the molar mass of X\(_2\)CO\(_3\)

Molar mass \(\frac{\text{Mass concentration in}gdm^{-3}}{\text{concentration in mol dm}^{-3}}\)

= \(\frac{10.6}{0.100}\) 

= 106 gmol\(^{-1}\)

 


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WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
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