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2010 WAEC Chemistry Theory (a) (i) Define a base according to Arrhenius concept. (ii) Give one example of an...

Chemistry
WAEC 2010

(a) (i) Define a base according to Arrhenius concept.

(ii) Give one example of an Arrhenius base.

(iii) Identify each of the following substances in aqueous solutions as strong electrolyte, non-electrolyte or weak electrolyte.

I. C\(_{12}\)H\(_{22}\)O\(_{11}\)  II. NH\(_3\) Ill. NaOH

(iv) Write a balanced equation to represent the reaction between CH\(_3\)COOH and KOH.

 

(b) Calculate the volume of 0.500 mol dm\(^{-3}\) HCI required to neutralize 20.00 cm\(^3\) of 0.300 mol dm\(^{-3}\) NaOH.

(c) Give the IUPAC name of each of the following salts: (ii) NaOCI; (iii) Mg(HCO\(_3\))\(_2\).

(d)(i) Define the term standard solution.

(ii) Consider the following compounds: NaOH and Na\(_2\)CO\(_3\). Which of the compounds is suitable for the preparation of a standard alkaline solution? Give reason for your answer

(iii) Fe completely reacted with dilute HCI.

I. Write an equation for the reaction

II. If 3.08g of Fe completely reacted with 50.0 cm\(^3\) of 2.20 mol dm\(^{-3}\) HCI, calculate the relative atomic mass of the metal.

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Explanation

(a)(i) A base is a substance which produces hydroxyl (OH\(^-\)) ions in its aqueous solution.

(ii) NaOH/Ca(OH)\(_2\)/ KOH.

(iii) I. non-electrolyte II. weak electrolyte HI. strong electrolyte.

(iv) CH\(_3\)COOH + KOH --> CH\(_3\)COOK + H\(_2\)O

 

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

V\(_A\) = \(\frac{C_BV_B}{C_A}\)

\(\frac{0.30 \times 20.50}{0.50}\) cm\(^3\)

 

(c)(i) sodium oxochiorate 1 (ii) magnesium  (iii) hydrogen trioxocarbonate(IV).

 

(d)(i) A standard solution is a solution whose concentration is (accurately) known.

(ii) Na\(_2\)CO\(_3\); Na\(_2\)CO\(_3\) is easy to weigh accurately because it is stable in air/non-deliquescent.

(iii) I. Fe\(_{(s)}\) + 2HCl\(_{(aq)}\) \(\to\) FeCl\(_{2(aq)}\) + H\(_{2(aq)}\).

II. number of moles of HCI used = 2.20 x \(\frac{50.0}{1000}\) = 0.11

mole ratio of HCI : Fe = 2 : 1

number of moles of Fe used = \(\frac{0.11}{2}\) =  0.055 mole
0.055 mole of Fe = 3.08g

0.1 mole of Fe = \(\frac{3.08 \times 1.0}{0.055}\) = 56g

Relative atomic mass of Fe = 56.


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