
Mercuric oxide, \[{\text{HgO}}\] , can be analysed by reaction with iodide and then titration with an acid. \[{\text{HgO + 4 }}{{\text{I}}^ - }{\text{ }} \to {\text{ HgI}}_4^{2 - }{\text{ + 2 O}}{{\text{H}}^ - }\] then, \[{\text{HgO}} \equiv 2{\text{O}}{{\text{H}}^ - } \equiv 2{{\text{H}}^ + }\] . So, if \[{\text{x moles}}\] \[{\text{HgO}}\] requires \[{\text{10 moles}}\] of a dibasic acid then what is the value \[x\] .
A) 40
B) 20
C) 5
D) 0
Answer
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Hint: A dibasic acid is an acid which gives two protons per molecule. One molecule of dibasic acid reaction reacts with two hydroxide ions.
Complete answer:
Mercuric oxide reacts with iodide ions to form the complex ion \[{\text{HgI}}_4^{2 - }\] and hydroxide ions. Write balanced chemical equation for the above reaction
\[{\text{HgO + 4 }}{{\text{I}}^ - }{\text{ }} \to {\text{ HgI}}_4^{2 - }{\text{ + 2 O}}{{\text{H}}^ - }\]
In this chemical equation, the atoms of mercury and iodine are balanced. But the atoms of oxygen and hydrogen are not balanced. The charges are balanced.
To balance the number of oxygen and hydrogen atoms, add a water molecule to the reactants side.
\[{\text{HgO + 4 }}{{\text{I}}^ - }{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{ HgI}}_4^{2 - }{\text{ + 2 O}}{{\text{H}}^ - }\]
This is the balanced chemical equation for the reaction
From this balanced chemical equation, We can conclude that \[{\text{1 mole HgO}} \equiv 2{\text{ mole O}}{{\text{H}}^ - }\]
We can titrate the liberated hydroxide ions with a dibasic acid such as sulphuric acid
\[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + 2 O}}{{\text{H}}^ - }{\text{ }} \to {\text{ 2}}{{\text{H}}_2}{\text{O + SO}}_4^{2 - }\]
From the above equation, We can conclude that \[{\text{1 mole }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_4} \equiv 2{\text{ mole O}}{{\text{H}}^ - }\]
Thus,
\[{\text{1 mole HgO}} \equiv 2{\text{ mole O}}{{\text{H}}^ - } \equiv {\text{1 mole }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_4}\]
or \[{\text{1 mole HgO}} \equiv {\text{1 mole }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_4}\]
Then, the number of moles of dibasic acid is equal to the number of moles of \[{\text{HgO}}\]
When \[{\text{x moles}}\] \[{\text{HgO}}\] are titrated with dibasic acid, \[{\text{x moles}}\] of dibasic acid are required.
But, \[{\text{x moles}}\] \[{\text{HgO}}\] requires 10 moles of a dibasic acid
Hence, \[x = 10\]
Hence, the correct answer is the option (D).
Note: In a titration reaction, the concentration of one solution is known. You can use this concentration to determine the concentration of another solution by using the formula \[{{\text{M}}_1} \times {{\text{V}}_1}{\text{ = }}{{\text{M}}_2} \times {{\text{V}}_2}\] . You can rearrange the formula as \[{{\text{M}}_2} = \dfrac{{{{\text{M}}_1} \times {{\text{V}}_1}{\text{ }}}}{{{{\text{V}}_2}}}\].
Complete answer:
Mercuric oxide reacts with iodide ions to form the complex ion \[{\text{HgI}}_4^{2 - }\] and hydroxide ions. Write balanced chemical equation for the above reaction
\[{\text{HgO + 4 }}{{\text{I}}^ - }{\text{ }} \to {\text{ HgI}}_4^{2 - }{\text{ + 2 O}}{{\text{H}}^ - }\]
In this chemical equation, the atoms of mercury and iodine are balanced. But the atoms of oxygen and hydrogen are not balanced. The charges are balanced.
To balance the number of oxygen and hydrogen atoms, add a water molecule to the reactants side.
\[{\text{HgO + 4 }}{{\text{I}}^ - }{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{ HgI}}_4^{2 - }{\text{ + 2 O}}{{\text{H}}^ - }\]
This is the balanced chemical equation for the reaction
From this balanced chemical equation, We can conclude that \[{\text{1 mole HgO}} \equiv 2{\text{ mole O}}{{\text{H}}^ - }\]
We can titrate the liberated hydroxide ions with a dibasic acid such as sulphuric acid
\[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + 2 O}}{{\text{H}}^ - }{\text{ }} \to {\text{ 2}}{{\text{H}}_2}{\text{O + SO}}_4^{2 - }\]
From the above equation, We can conclude that \[{\text{1 mole }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_4} \equiv 2{\text{ mole O}}{{\text{H}}^ - }\]
Thus,
\[{\text{1 mole HgO}} \equiv 2{\text{ mole O}}{{\text{H}}^ - } \equiv {\text{1 mole }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_4}\]
or \[{\text{1 mole HgO}} \equiv {\text{1 mole }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_4}\]
Then, the number of moles of dibasic acid is equal to the number of moles of \[{\text{HgO}}\]
When \[{\text{x moles}}\] \[{\text{HgO}}\] are titrated with dibasic acid, \[{\text{x moles}}\] of dibasic acid are required.
But, \[{\text{x moles}}\] \[{\text{HgO}}\] requires 10 moles of a dibasic acid
Hence, \[x = 10\]
Hence, the correct answer is the option (D).
Note: In a titration reaction, the concentration of one solution is known. You can use this concentration to determine the concentration of another solution by using the formula \[{{\text{M}}_1} \times {{\text{V}}_1}{\text{ = }}{{\text{M}}_2} \times {{\text{V}}_2}\] . You can rearrange the formula as \[{{\text{M}}_2} = \dfrac{{{{\text{M}}_1} \times {{\text{V}}_1}{\text{ }}}}{{{{\text{V}}_2}}}\].
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