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Chemistry

Using dilute sulphuric acid how would you differentiate between :

(a) Copper and magnesium

(b) Sodium carbonate, sodium sulphide and sodium sulphite

How would you identify the gaseous product evolved.

Practical Chemistry

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Answer

(a) Magnesium on reaction with dil. sulphuric acid produces a colourless, odourless gas with brisk effervescence.

The gas evolved is hydrogen as it burns with a pale blue flame producing a pop sound.

Mg+H2SO4 [dil.]MgSO4+H2\text{Mg} + \underset{\text{ [dil.]}}{\text{H}2\text{SO}4}\longrightarrow \text{MgSO}4+ \text{H}2

Copper does not react with dil. sulphuric acid liberating hydrogen as it is lower in metal reactivity series than hydrogen.

(b) Sodium carbonate — When dil. sulphuric acid is added to sodium carbonate, the gas evolved turns lime water milky but has no effect on potassium permanganate solution or potassium dichromate solution. This confirms that the gas evolved is Carbon dioxide.

Na2CO3+H2SO4 [dil.]Na2SO4+H2O+CO2\text{Na}2\text{CO}3 + \underset{\text{ [dil.]}}{\text{H}2\text{SO}4}\longrightarrow \text{Na}2\text{SO}4+ \text{H}2\text{O} + \text{CO}2

2NaHCO3+H2SO4 [dil.]Na2SO4+2H2O+2CO22\text{NaHCO}3 + \underset{\text{ [dil.]}}{\text{H}2\text{SO}4}\longrightarrow \text{Na}2\text{SO}4+ 2\text{H}2\text{O} + 2\text{CO}_2

Sodium sulphide — When dil. sulphuric acid is added to sodium sulphide, colourless gas is evolved with a rotten egg smell that turns moist lead acetate paper silvery black. This confirms that the gas evolved is hydrogen sulphide.

Na2S+H2SO4 [dil.]Na2SO4+H2S\text{Na}2\text{S} + \underset{\text{ [dil.]}}{\text{H}2\text{SO}4}\longrightarrow \text{Na}2\text{SO}4+ \text{H}2\text{S}

Sodium sulphite — When dil. sulphuric acid is added to sodium sulphite, colourless gas is evolved with a suffocating odour. It turns lime water milky and pink potassium permanganate solution colourless. It also turns orange potassium dichromate solution clear green, confirming the presence of sulphur dioxide.

Na2SO3+H2SO4 [dil.]Na2SO4+H2O+SO2\text{Na}2\text{SO}3 + \underset{\text{ [dil.]}}{\text{H}2\text{SO}4}\longrightarrow \text{Na}2\text{SO}4+ \text{H}2\text{O} + \text{SO}2

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