(a) In an experiment, 25.0 cm\(^{3}\) of H\(_{2}\)SO\(_{4}\) completely neutralized 24.0 cm\(^{3}\) of a 0.1 50 mol dm\(^{-3}\) aqueous KOH using a suitable indicator.
(i) Write a balanced chemical equation for the reaction.
(ii) Calculate the concentration of the acid solution.
(b)i. A burning magnesium ribbon was placed in a gas jar containing Carbon (IV) oxide.
- Write an equation for the reaction.
- Explain briefly why the magnesium ribbon burns in carbon(lV) oxide although the gas does not support combustion.
- Calculate the percentage mass of nitrogen in magnesium trioxonitrate (V). |N = 140. O = 16.0. Mg = 24.01]
(c) Consider the following organic compound: CH\(_{3}\)CH\(_{2}\)CH = CHCOOH.
(i) State two chemical reactions which could he sed te identify the compound.
(ii) What would be observed in each of the reactions stated in c(i)
(d) Describe briefly how soap is manufactured using pellets of sodium hydroxide and vegetable oil
(e) Define the term electronegativity
(a)(i) Equation for the neutralization
\[H_2SO_{4(aq)} + 2KOH_{(aq)} \to K_2SO_{4(aq)} + 2H_2O_{(l)}\]
(a)(ii) Concentration of the acid
Moles of KOH \(= \dfrac{24.0}{1000} \times 0.150 = 3.6 \times 10^{-3}\,mol\).
From the equation, 2 mol KOH react with 1 mol \(H_2SO_4\), so moles of acid \(= \dfrac{3.6 \times 10^{-3}}{2} = 1.8 \times 10^{-3}\,mol\).
\[[H_2SO_4] = \frac{1.8 \times 10^{-3}}{25.0/1000} = 0.072\,mol\,dm^{-3}\]
(b) Magnesium burning in carbon(IV) oxide
Equation:
\[2Mg_{(s)} + CO_{2(g)} \to 2MgO_{(s)} + C_{(s)}\]
Explanation: magnesium is a very reactive metal and a powerful reducing agent. The heat of the burning ribbon is enough to decompose carbon(IV) oxide; magnesium removes the oxygen from the \(CO_2\), reducing it to carbon (seen as black specks) while itself being oxidized to white magnesium oxide. So the ribbon continues to burn even though \(CO_2\) does not normally support combustion.
Percentage mass of nitrogen in magnesium trioxonitrate(V), \(Mg(NO_3)_2\):
\[M = 24 + 2(14 + 48) = 148\]\[\%N = \frac{2 \times 14}{148} \times 100 = 18.92\%\]
(c) The compound \(CH_3CH_2CH=CHCOOH\)
This compound has both a carbon-carbon double bond (unsaturation) and a carboxylic acid (\(-COOH\)) group.
- (i) Reaction 1: add bromine water. Reaction 2: add sodium trioxocarbonate(IV) (\(Na_2CO_3\)) solution.
- (ii) With bromine water: the red-brown bromine water is decolourized (turns colourless), confirming the \(C=C\) double bond. With \(Na_2CO_3\): brisk effervescence occurs, giving off a colourless gas (\(CO_2\)) that turns limewater milky, confirming the acidic \(-COOH\) group.
(d) Manufacture of soap
Vegetable oil is boiled with concentrated sodium hydroxide (caustic soda) solution. The oil is hydrolyzed (saponified) to give soap (the sodium salt of the fatty acid) and glycerol. Common salt (sodium chloride) is then added to the mixture to salt out (precipitate) the soap, which floats to the top; it is filtered off, pressed and moulded.
(e) Electronegativity
Electronegativity is the relative tendency (power) of an atom to attract the shared pair of electrons in a covalent bond towards itself.
(a)(i) Equation for the neutralization
\[H_2SO_{4(aq)} + 2KOH_{(aq)} \to K_2SO_{4(aq)} + 2H_2O_{(l)}\]
(a)(ii) Concentration of the acid
Moles of KOH \(= \dfrac{24.0}{1000} \times 0.150 = 3.6 \times 10^{-3}\,mol\).
From the equation, 2 mol KOH react with 1 mol \(H_2SO_4\), so moles of acid \(= \dfrac{3.6 \times 10^{-3}}{2} = 1.8 \times 10^{-3}\,mol\).
\[[H_2SO_4] = \frac{1.8 \times 10^{-3}}{25.0/1000} = 0.072\,mol\,dm^{-3}\]
(b) Magnesium burning in carbon(IV) oxide
Equation:
\[2Mg_{(s)} + CO_{2(g)} \to 2MgO_{(s)} + C_{(s)}\]
Explanation: magnesium is a very reactive metal and a powerful reducing agent. The heat of the burning ribbon is enough to decompose carbon(IV) oxide; magnesium removes the oxygen from the \(CO_2\), reducing it to carbon (seen as black specks) while itself being oxidized to white magnesium oxide. So the ribbon continues to burn even though \(CO_2\) does not normally support combustion.
Percentage mass of nitrogen in magnesium trioxonitrate(V), \(Mg(NO_3)_2\):
\[M = 24 + 2(14 + 48) = 148\]\[\%N = \frac{2 \times 14}{148} \times 100 = 18.92\%\]
(c) The compound \(CH_3CH_2CH=CHCOOH\)
This compound has both a carbon-carbon double bond (unsaturation) and a carboxylic acid (\(-COOH\)) group.
- (i) Reaction 1: add bromine water. Reaction 2: add sodium trioxocarbonate(IV) (\(Na_2CO_3\)) solution.
- (ii) With bromine water: the red-brown bromine water is decolourized (turns colourless), confirming the \(C=C\) double bond. With \(Na_2CO_3\): brisk effervescence occurs, giving off a colourless gas (\(CO_2\)) that turns limewater milky, confirming the acidic \(-COOH\) group.
(d) Manufacture of soap
Vegetable oil is boiled with concentrated sodium hydroxide (caustic soda) solution. The oil is hydrolyzed (saponified) to give soap (the sodium salt of the fatty acid) and glycerol. Common salt (sodium chloride) is then added to the mixture to salt out (precipitate) the soap, which floats to the top; it is filtered off, pressed and moulded.
(e) Electronegativity
Electronegativity is the relative tendency (power) of an atom to attract the shared pair of electrons in a covalent bond towards itself.
