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# A quantity of 0.62gm of $N{a_2}C{O_3}.{H_2}O$ is added to $100{\text{ }}ml$ of $0.1N - {H_2}S{O_4}$ solution. The resulting solution would beA) AcidicB) AlkalineC) NeutralD) Buffer

Last updated date: 14th Jun 2024
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Hint: The nature of solution depends upon the component which has a higher number of equivalents
-That means in a Solution of a alkali and acid if the acid has more number of equivalents the nature of solution will be acidic else if the alkali have more number of equivalents than the solution will be alkaline in nature
-Finally, if both of them have equal number of equivalents then the solution will be neutral in nature
Formula Used:
$Number\,of\,equivalents = \dfrac{{Given\,\,weight}}{{Equivalent\,Weight}}$
$Number\,of\,equivalents = Normality \times Volume\,in\,l$

Complete step by step solution:
The reaction occurring here is:
$N{a_2}C{O_3}.{H_2}O + {H_2}S{O_4} \to N{a_2}S{O_4} + {H_2}C{O_3} + {H_2}O$
As in the given question the molecular weight of $N{a_2}C{O_3}.{H_2}O$
Will be $M = 23 \times 2 + 12 + 3 \times 16 + 18$
$M = 142$
But for the given question we need equivalent weight which for an ionic compound is given as follow
$E = \dfrac{{Molecular\,Weight}}{{valency}}$
Hence equivalent weight of $N{a_2}C{O_3}.{H_2}O$
Will be ${E_1} = \dfrac{{124}}{2}$
${E_1} = 62$

Also given weight of $N{a_2}C{O_3}.{H_2}O$ is $0.62gm$
Number of equivalents of $N{a_2}C{O_3}.{H_2}O$ $= \dfrac{{Given\,\,weight}}{{Equivalent\,Weight}}$
$= \dfrac{{0.62}}{{62}}$
$= 0.01$
Now equivalents of $0.1N - {H_2}S{O_4}$ can be calculated using the formula
$Number\,of\,equivalents = Normality \times Volume$
Converting the volume of $0.1N - {H_2}S{O_4}$ $100ml$ into litre and using it as follows
We get equivalents of $0.1N - {H_2}S{O_4}$ $= 0.1 \times 100 \times {10^{ - 3}}$
$= 0.01$
-As we can see that the number of equivalents of $100{\text{ }}ml$ of $0.1N - {H_2}S{O_4}$ comes out to be equal to number of equivalents of $0.62gm$ of $N{a_2}C{O_3}.{H_2}O$ we can conclude that the resulting solution would be neutral in nature hence the option ‘C’ is the correct solution for the given question.

-A buffer solution resists change in its $pH$
-$N{a_2}C{O_3}$ is a metal carbonate and in the given question it consist two water molecules as water of crystallization but that does not affect its strong alkaline nature which was the reason it could neutralize
${H_2}S{O_4}$ Which is a very strong acid