Question

The formula weight of an acid is $82$. In a titration, $100c{m^3}$of a solution of this acid containing $39.0g$of the acid per litre were completely neutralised by $95.0c{m^3}$of aqueous $NaOH$ containing $40.0g$of $NaOH$ per litre. What is the basicity of the acid?

Answer 1

Hint:Given that the solution is completely neutralized, from this we should understand the milliequivalent of acid and the milliequivalent of base will be equal. Solving this we should get our basicity of an acid.


Complete solution:
We have asked to find the basicity of an acid, for that we have given a titration which means a solution of acid is being neutralized by the base. Here the acid is not specified and have to find its basicity.
Basicity of acid is generally said to be the number of hydrogen ions which can be produced by one molecule of the acid. In volumetric analysis of acid-bas titration we have given an acid of weight $82g$.
Weight of acid = $82g$
The equivalent weight of an acid will be given as molecular weight by its valency factor. This valency factor is the basicity of that acid. To find out that let us use normality of the solution.
$Equivalent\;weight\; = \dfrac{{\;82}}{n}$
Where $n$ is the basicity of acid.

We know that normality is the number of gram equivalent of solute dissolver per litre of the solution.
$Normality,\;N = \dfrac{{weight\;of\;solute}}{{Equivalent\;weight}} \times \dfrac{{1000}}{{volume\;of\;solution\left( {ml} \right)}}$
Substituting the above values we get.

Weight of acid $ = 39g{L^{ - 1}}$

$N = \dfrac{{39}}{{\left( {\dfrac{{82}}{n}} \right)}} \times \dfrac{{1000}}{{1000}}$, since weight is given per litre.

Now we have solution of base which is$NaOH$, where,$40g$ of $NaOH$ is present in per litre given in question which makes the normality of $NaOH$as $1$. Mathematically,
$Normality,\;N = \dfrac{{40}}{{40}} \times \dfrac{{1000}}{{1000}}$

Since the weight of solute given in question is $40g$ and the equivalent weight of $NaOH$is also $40g$. Since the weight given is in per litre we take the volume of solution as $1000ml$. So we got the normality as $1$.

Here, $Eq.\;weight\; = \;Normality \times Volume$

Also,$Eq.\;weight\;of\;acid\; = \;Eq.\;weight\;of\;base$
Now substituting the values
Given, Volume of acid = $100c{m^3}$=$100ml$ and volume of base = $95c{m^3}$=$95ml$

$\dfrac{{39}}{{\left( {\dfrac{{82}}{n}} \right)}} \times 100 = 1 \times 95$

Solving this,
$\dfrac{{39{\text{n}}}}{{82}} = \dfrac{{95}}{{100}}$
$n = \dfrac{{95 \times 82}}{{100 \times 39}}$
i.e., $n = 1.99$

Approximating this $2$, we get our acid as a dibasic.


Note:Here, we used normality to express the concentration of solution since we know that equivalent weight of an acid and a base will be equal and since normality is also expressed in terms of equivalent weight we choose normality for this question.