Question

How much amount of $CuS{O_4}.5{H_2}O$ is required for liberation of 2.54g of ${I_2}$ when titrated with KI ?
A) 2.5g
B) 4.99g
C) 2.4g
D) 1.2g

Answer 1

The number of moles required to liberate ${I_2}$ when mixed with $CuS{O_4}.5{H_2}O$ when titrated with KI is calculated with the help of molecular weight of the compound involved in the titration reaction.

Complete step by step answer:
The titration reaction is a technique to determine concentration of an analyte when mixed in a reaction. Volume of the titrants are added to the analyte and indicator until the indicator changes color in reaction to the titrant saturation threshold, which represents that the amount of titrant is balanced off with the amount of analyte present
It is a type of acid base titration in which neutralization between an acid and a base takes place in the presence of an indicator when mixed in solution. The acid–base indicator indicates the endpoint of the titration by changing color.
When ${I_2}$ is titrated with $KI$ in presence of an oxidizing agent $CuS{O_4}.5{H_2}O$. The amount of iodine liberated from iodide is equivalent to the quantity of the oxidizing agent present in the reaction.
The reaction takes place as:
$2CuS{O_4}.5{H_2}O + 4KI \to C{u_2}{I_2} + 2{K_2}S{O_4} + {I_2} + 10{H_2}O$
Molecular weight of ${I_2}$ is $254g$
The molecular weight of $2CuS{O_4}.5{H_2}O$
= [2(63.5 + 32 + 64) + 10(18)]
= 499g
254g of ${I_2}$ is liberated by 499g of $CuS{O_4}.5{H_2}O$
To calculate the 2.54g of ${I_2}$is liberated by xg $CuS{O_4}.5{H_2}O$
$x = \dfrac{{499}}{{254}}X2.54 = 4.99g$

Hence the correct answer is (B).

Note: This titration reaction takes place in the presence of starch as an indicator to give a coloured complex when being titrated. The concentration of a titrant and an analyte is measured by Henderson-Hasselbalch equation.