Question

What weight of $N{a_2}C{O_3}$ of $85\%$ purity would be required to prepare $45.6{\text{ ml}}$ of $0.235{\text{ M}}$ ${H_2}S{O_4}$ ?

Answer 1

To proceed with this question, we will apply the rule of neutralization reaction. So basically in this question it is asked that what weight of $N{a_2}C{O_3}$ of $85\%$ purity would be required to neutralize $45.6{\text{ ml}}$ of $0.235{\text{ M}}$ ${H_2}S{O_4}$ .

Complete answer:
Since, because of the neutralization reaction we can say that:
Milliequivalents of $N{a_2}C{O_3}$ will be equal to the milliequivalents of the ${H_2}S{O_4}$ .
Hence , we can write:
Meq of $N{a_2}C{O_3}$ $=$ Meq of ${H_2}S{O_4}$ , For complete neutralization reaction.
Now, Meq of $N{a_2}C{O_3}$ $=$ $0.235 \times 45.6$
Also by the formula , we can write the Meq of $N{a_2}C{O_3}$ in other way as:
$\dfrac{w}{{equivalent{\text{ }}weight}} \times 1000 = 45.6 \times 0.235$
So, let's find out the equivalent weight of $N{a_2}C{O_3}$ .
Also , note that here we have used $1000$ in the equation because everything is in milliequivalents
So, our equation becomes:
$\dfrac{w}{{\dfrac{{106}}{2}}} \times 1000 = 45.6 \times 0.235$
$w = 0.5679{\text{ g}}$
Hence our calculated weight of $N{a_2}C{O_3}$ is , $w = 0.5679{\text{ g}}$
Now, moving forward , it is given in the question that $N{a_2}C{O_3}$ is $85\%$ pure.
It means that,
$85$ gram $N{a_2}C{O_3}$ will be present in $100$ gram of sample
So, $1$ gram of $N{a_2}C{O_3}$ will be present in $= \dfrac{{100}}{{85}}$ gram of sample
And $0.5679{\text{ }}$ gram of $N{a_2}C{O_3}$ will be present in $= \dfrac{{100}}{{85}} \times 0.5679$
$= 0.668$ gram
Therefore the correct answer is $= 0.668$ gram
So, $0.668$ $N{a_2}C{O_3}$ of $85\%$ purity would be required to prepare $45.6{\text{ ml}}$ of $0.235{\text{ M}}$ ${H_2}S{O_4}$ .

Note:
Miliequivalent is one- thousandth of the equivalent of the chemical species , compound , element or radical. An equivalent is the amount of substance that reacts with an arbitrary amount of another substance in a particular chemical reaction. It is a unit of measurement of the compounds that are thoroughly used in the chemistry and the biological sciences. The mass of an equivalent is called its equivalent weight.