Question

If $5g$ Of ${K_2}S{O_4}$ were dissolved in $250ml$ of solution. How many ml of this solution should be used so that $1.2g$ of $BaS{O_4}$ may be precipitated from $BaC{l_2}$ solution?

Answer 1

Hint: We know that the number of equivalents per mole of an ion equals the charge on the ion and this leads to the definition of milliequivalent. Equivalent per liter is a unit of concentration$mEq/L.$
Mathematically, milliequivalent is represented as,
$mEq = \dfrac{{\left( {Mass} \right)\left( {Volume} \right)}}{{{\text{Molecular weight}}}}$
Example: To find the milliequivalent of potassium in $750ml$ of solution contains $58.65mg/L$ of potassium ion and the valence of potassium is $1$.
Using the above equation,
$mEq = \dfrac{{\left( {58.65} \right)\left( 1 \right)}}{{39.1}} = 1.5mEq$
The milliequivalent can likewise be determined utilizing the formula,
$mEq = Concentration \times volume$

Complete step by step answer:
We can write the balanced equation for this reaction as,
$BaC{l_2} + {K_2}S{O_4} \to BaS{O_4} + 2KCl$
Given data contains,
The amount of potassium sulphate is$5g$.
The amount of barium sulphate is $1.2g$.
The volume of potassium sulphate is $250ml$.
The molecular weight of potassium sulphate is $174g$.
The molecular weight of barium sulphate is $233g$.
The milliequivalent of barium sulphate formed$ = \dfrac{{1.2}}{{\dfrac{{233}}{2}}} \times 1000 = 10.30$
The milliequivalent of potassium sulphate formed$ = \dfrac{{5 \times 1000}}{{\dfrac{{174}}{2} \times 250}}$
The volume of potassium sulphate can calculated as,
$mEq = Concentration \times volume$
$10.30 = = \dfrac{{5 \times 1000}}{{\dfrac{{174}}{2} \times 250}} \times V$
On simplifying we get,
$ \Rightarrow V = 44.8ml$

Thus, the volume of potassium sulphate $ = 44.8ml$.

Note: We can likewise characterize the equivalent weight of the substance as the proportion of the sub-atomic weight or mass of the compound to the n-factor or the acidity or basicity. We can figure the n-factor by deciding the adjustment in oxidation state. The 'n' factor of an acid is the quantity of particles replaced by one mole of acid. The n-factor for acid isn't the quantity of moles of usable Hydrogen molecules present in one mole of acid. The oxidation condition of the single component is zero however on the off chance that it has any charge present on it, at that point it is considered as n-factor while basicity for the acidic substance and the acidity is characterized for the essential substance.
Example:
The equivalent weight of the sulfur dioxide is calculated as,
We know the molecular mass of the sulfur dioxide is $64g/mol$
Now, we calculate the n-factor. First, calculate the change in the oxidation state of sulfur in the reactions.
The given reactions is,
$S{O_2} + 2{H_2}S = 3S + 2{H_2}O$
The oxidation state of sulfur is changed from $ + 4$ to zero. Thus the n-factor is $4.$
The equivalent weight of sulfur $ = \dfrac{{64}}{4} = 16g$
$5S{O_2} + 2KMn{O_4} + 2{H_2}O = {K_2}S{O_4} + 2MnS{O_4} + 2{H_2}S{O_4}$
The oxidation state of sulfur is changed from $ + 4$ to \[ + 6\]. Thus the n-factor is $2$.
The equivalent weight of sulfur $ = \dfrac{{64}}{2} = 32g$
Thus, the equivalent weight of sulfur in reaction a is $16g$ and the equivalent weight of sulfur in reaction a is $32g$