Question

$3.0g$ sample of $KOCl$ and $CaOC{l_2}$ is dissolved in water to prepare $100mL$ solution which requires $100mL$ of $0.15M$ acidified ${K_2}{C_2}{O_4}$ for the end point. The clear solution is now treated with excess of $AgN{O_3}$ solution which precipitates $2.87g$ of $AgCl.$ The mass percentage of $KOCl$ and $CaOC{l_2}$ in the mixture is :

Answer 1

In titration, mass equivalent of one substance becomes equal to the mass equivalent of another substance. Titration is used to find the unknown concentration of substance by treating it with substance of known strength and molarity.

Complete step by step answer: As we know that titration is a common laboratory method of quantitative chemical analysis to determine the concentration of an analyte.
Now according to the question, a mixture of $KOCl$ and $CaOC{l_2}$when titrated, they give the chloride salt. The redox reactions are written below. The formation of $C{l^ - }$ from $KOCl$ is written as
$Cl\;{O^ - } + 2{e^ - }\; \to \;C{l^ - }$
$\left( {x = + 1} \right)\left( {x = - 1} \right)$
Here x represents the oxidation state of $Cl$ on both reactant and product side. The oxidation changes from $+ 1$ to $- 1$ , so the change in oxidation state (n) is $\left| { + 1( - \left( { - 1} \right)} \right| = 2,$
The formation of $C{l^ - }$ from $CaOC{l_2}$ is written as
$C{l_2}{O^{2 - }}\;\;\;\; + \;2{e^ - }\;\; \to \;\;\;C{l^ - }$
$\left( {x = 0} \right)\left( {x = - 1} \right)$
The change in oxidation state is (n) is $1$ .
Now conversion of $C{r_2}{O_4}^{2 - }$ takes place as
$C{r_2}{O_4}^{2 - } \to \;\;2C{O_2} + \;2{e^ - }$
$\left[ {x = 3} \right]\;\;\left[ {x = 4} \right]$
Here change in oxidation state is $= 2$
Now let a and b millimoles of $CaOC{l_2}$ are present in the mixture respectively.
Hence, according to question
Mass equivalence of $KOCl$ $+$ Mass Equivalence of $CaOC{l_2}$= Mass Equivalence of ${K_2}C{r_2}{O_4}$
$\Rightarrow$ $2a + 2b = 100 \times 0.15 \times 2$
$\Rightarrow$ $2a + 2b\; = \;30\;\;\;\; - - - \left( 1 \right)$
[We have multiplied a and ${K_2}C{r_2}{O_4}$ because their change in oxidation state is $2$ and with b because there are $2\;C{l^ - }$ atoms present in its compound which changes its oxidation state]
Also,
Millimoles of $C{l^ - }$ from $KOCl$ $+$ Millimoles of $C{l^ - }$ from $CaOC{l_2}$ = Millimoles of $AgCl$
$1 \times a$ [ because only on $C{l^ - }$ present ] $+ 2 \times b$ [ because two $C{l^ - }$ present ]$= \dfrac{{Given{\text{ mass of AgCl}}}}{{Molar{\text{ mass of AgCl}}}} \times {10^3}$
$\Rightarrow$ $a + 2b\; = \;\dfrac{{2.87}}{{143.5}} \times {10^3}$
$\Rightarrow$ $a + 2b\; = 20\;\; - - \left( 2 \right)$
[because mass of $AgCl$ given is $2.87g$ & Molar Mass of $AgCl = 143.5g$ ]
On Comparing and Equating $\left( 1 \right)$ & $\left( 2 \right)$, we got $a = 10$ and $b = 5$
Now total weight of given sample is $3g.$
Millimoles of $KOCl\, = 10$
Millimoles of $CaOC{l_2} = 5$
$\therefore \;\%$ of $KOCl\; =$ $\dfrac{{\operatorname{No} .\;\operatorname{moles} \,of\;KOCl}}{{\operatorname{No} .\;mole\;of\;KOCl\;\operatorname{given} }} \times 100\;\; - - - - \left( 3 \right)$
$\Rightarrow$ $\dfrac{{10 \times {{10}^{ - 3}}}}{3} \times 90.5 \times 100 = 30.1\%$
[Because molecular weight is $90.5g$ ]
$\therefore \% \;$ of $CaOC{l_2}$ = $\dfrac{{5 \times {{10}^{ - 3}}12}}{3} \times 100 = 21.2\%$
[using Equation $3$ and we know Mol. wt. of $CaOC{l_2}$ is $127$ ]

Note: In this question, we have used the concept of oxidation state also. Oxidation state is defined as the valency or power shown by an element when bonded with other atoms. Oxidation state can be calculated by comparing variances of bonded atoms.