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17 questions · timed · auto-graded

Question 12 Marks
Distinguish between the following pair of compound using the test given within bracket.
A lead salt and a zinc salt (using excess ammonium hydroxide
Answer
$Pb\left(NO_3\right)_2+2 NH_4 OH \rightarrow Pb(OH)_2+2 NH_4 NO_3$
On adding excess of $NH _4 OH$, chalky white ppt. of insoluble $Pb ( OH )_2$ is formed.
$ZnSO_4+2 NH_4 OH \rightarrow Zn(OH)_2+\left(NH_4\right)_2 SO_4$
With excess of $NH _4 OH$, white gelatinous ppt. of soluble $Zn ( OH )_2$ is formed.
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Question 22 Marks
Distinguish between the following pair of compound using the test given within bracket.
Iron (II) Sulpate an and iron(III) using ammonium hydroxide)
Answer
Iron (II):
$
\underset{\text { (Light green solution) }}{ FeSO _4}+2 NH _4 OH \longrightarrow \underset{\text { (Dirty green ppt.) }}{ Fe ( OH )_2}+\left( NH _4\right)_2 SO _4
$Iron (III):
$
\underset{\text { (Yellow solution) }}{ FeCl _3}+3 NH _4 OH \longrightarrow \underset{\text { (Reddish brown ppt.) }}{ Fe ( OH )_3}+ NH _4 Cl
$
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Question 32 Marks
State two relevant observation of the following:
Ammonium hydroxide solution is added to zinc nitrate solution in minimum quantities and then in excess
Answer
(a) When $NH_4OH$ is added to zinc nitrate solution in minimum quantity, it forms a gelatinous white precipitate.

(b) When added in excess, it dissolves to form a complex salt.
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Question 42 Marks
State two relevant observation of the following:
Ammonium hydroxide solution is added to copper (II) nitrate solution in small quantities and then in excess
Answer
(a) When $NH_4OH$ is added to copper (II) nitrate solution in small quantities, a pale blue precipitate is observed.
(b) When added in excess, $NH_4OH$ dissolves to give an inky blue solution forming a complex salt.
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Question 52 Marks
One chemical test that would enable you to distinguish between the following pair of chemicals. Describe what happens with each chemical or state no visible reaction.
Sodium sulphate solution and sodium chloride solution.
Answer
Sodium sulphate solution Sodium chloride solution
When Sodium sulphate solution is treated with Barium chloride solution, a white precipitate is formed which is insoluble in all the mineral acids.When Sodium chloride solution is treated with Barium chloride solution, no visible reaction is observed.
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Question 62 Marks
Sodium hydroxide solution can be used to distinguish between i. iron (II) sulphate solution and (ii) iron (III) sulphate solution; because these solutions give different coloured precipitates with sodium hydroxide solution. Give the colour of the precipitate formed with each of the solution.
Answer
Sodium hydroxide solution gives dirty green coloured precipitates with iron (II)sulphate solution.
With iron (III) sulphate solution sodium hydroxide solution gives reddish brown precipitates.
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Question 72 Marks
What will you observe when barium chloride solution is added to iron(II) sulphate solution.
Answer
When barium chloride solution is added to iron (II) sulphate solution it gives white precipitate of $BaSO_4$​​​​​​​.
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Question 82 Marks
Write equation for the reaction that will take place when copper sulphate solution is added to sodium hydroxide solution.
Answer
For the reaction that will take place when copper sulphate solution is added to sodium hydroxide solution the equation is as:
$
CuSO _4+2 NaOH \longrightarrow Cu ( OH )_2+ Na _2 SO _4
$
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Question 92 Marks
Salts of [normal / transition] elements are generally coloured. From the ions $K ^{1+}, Cr ^{3+}$, $Fe ^{2+}, Ca ^{2+}, SO _3^{2-}, Mn _4^{1-}, NO _3^1$ the ions generally coloured are
Answer
Salts of transition elements are generally coloured. From the ions $K ^{1+}, Cr ^{3+}, Fe ^{2+}, Ca ^{2+}$,
$
SO _3^{2-}
$
$
Mn _4^{1-}
$
$
NO _3^1
$
the ions generally coloured are $Cr ^{3+}, Fe ^{2+}, MnO _4{ }^{4-}$.
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Question 102 Marks
Name a solution that will separate the component of the following mixture: CuO from ZnO.
Answer
CuO from ZnO:
Sodium hydroxide solution can separate CuO from ZnO as CuO precipitates remains insoluble in excess of NaOH solution while ZnO precipitates are soluble in excess of NaOH solution .
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Question 112 Marks
Name a solution that will separate the component of the following mixture: CaO from PbO.
Answer
CaO from PbO:
Sodium hydroxide solution can separate CaO from PbO as CaO precipitates are sparingly soluble in excess of NaOH solution while PbO precipitates are soluble in excess of NaOH solution.
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Question 122 Marks
Name a solution that will separate the component of the following mixture: $Zn(OH)_2$ from $Pb(OH)_2$
Answer
$Zn(OH)_2$ from $Pb(OH)_2$ :
Ammonium hydroxide $\left( NH _4 OH \right)$ solution can separate
$Zn ( OH )_2$ from $Pb ( OH )_2$ as $Zn ( OH )_2$ precipitates are dissolved in excess of $NH _4 OH$ solution while $Pb ( OH )_2$ precipitates are insoluble in excess of $NH _4 OH$ solution.
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Question 132 Marks
What do you observe when caustic soda solution is added to the following solution: first a little and then in excess : $CuSO_4$
Answer
$CuSO _4+2 NaOH \longrightarrow Cu ( OH )_2 \downarrow+ Na _2 SO _4$
Pale blue ppt.
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Question 142 Marks
What do you observe when caustic soda solution is added to the following solution: first a little and then in excess : $Pb(NO_3)_2$
Answer
$Pb ( NO )_3+2 NaOH \longrightarrow Pb ( OH )_{\text {White ppt. }}^{\downarrow}+2 NaNO _3$
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Question 152 Marks
What do you observe when caustic soda solution is added to the following solution: first a little and then in excess : $ZnSO_4$
Answer
$ZnSO _4+2 NaOH \longrightarrow \underset{\text { White gelatinous ppt. }}{ Zn ( OH )_2 \downarrow}+ Na _2 SO _4$
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Question 162 Marks
What do you observe when caustic soda solution is added to the following solution: first a little and then in excess : $FeCl_3$
Answer
$FeCl _3+3 NaOH \longrightarrow \underset{\text { Reddish Brown ppt. }}{ Fe ( OH )_3 \downarrow}+3 NaCI$
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Question 172 Marks
Using ammonium hydroxide, how would you distinguish between the following pair of ions in solution?
Ferrous ion and Ferric ion.
Answer
Ferrous ion and Ferric ion
(i)
$
FeSO _4+2 NH _4 OH \longrightarrow Fe ( OH )_2 \downarrow+\left( NH _4\right)_2 SO _4
$
$Fe ( OH )_2$ forms dirty green precipitates.
(ii)
$
Fe _2\left( SO _4\right)_3+6 NH _4 OH \longrightarrow 2 Fe ( OH )_3 \downarrow+3\left( NH _4\right)_2 SO _4
$
$Fe ( OH )_3$ forms reddish brown precipitates.
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