Answer
Verified
42.6k+ views
Hint: A solution is formed by the combination of solute and solvent. In the given question, impure solution of sodium chloride means solvent mixed with weak electrolyte, sodium chloride forming an impure solution or can say sodium chloride is not completely dissolved or dissociates into ions like \[N{{a}^{+}}\] and \[C{{l}^{-}}\] in solvent and making solution impure. We can purify it when it is completely non-electrolyte and change it to its NaCl, crystal form and we can collect it without any solvent.
Complete step by step solution:
We can get pure solvent and solute, sodium chloride when dissolved part of sodium chloride in the solvent gets completely undissolved or can say it completely precipitates on the bottom of the container. This can be possible if we hindered the ionisation of sodium chloride through other substances.
That substance is hydrogen chloride gas which helps to stop the ionisation of sodium chloride. Hydrogen chloride is a very good electrolyte (a good electrolyte can be dissociated into ions easily), act like two electrodes in the solution, one electrode is an anode (positively charged) that is \[{{H}^{+}}\] and another is a cathode (negatively charged) that is \[C{{l}^{-}}\].
On the other hand, as sodium chloride which is a weak electrolyte, gets partially ionised as \[N{{a}^{+}}\] and \[C{{l}^{-}}\] and is already present in the solution so, as soon as hydrogen chloride gas is passed and HCl gets ionised, chloride ion concentration in the solution will increase and thus, repulsion created between chloride ions. Due to this repulsion (common ion effect), mostly all chlorine forms bonds with sodium ions again present in the solution, and the solution is left with no more ionised form of sodium chloride. In this way, we can purify sodium chloride.
Thus, the correct option is B.
Note: Buffer solution is that solution whose pH remains constant even if we add even small amount of acid and base (do not allow any change) but in this question, we need to change the impure form of sodium chloride to pure form so, that buffer action does not help to purify the substance. Also, hydrolysis of salt gives a new product (NaOH) instead of pure sodium chloride.
Complete step by step solution:
We can get pure solvent and solute, sodium chloride when dissolved part of sodium chloride in the solvent gets completely undissolved or can say it completely precipitates on the bottom of the container. This can be possible if we hindered the ionisation of sodium chloride through other substances.
That substance is hydrogen chloride gas which helps to stop the ionisation of sodium chloride. Hydrogen chloride is a very good electrolyte (a good electrolyte can be dissociated into ions easily), act like two electrodes in the solution, one electrode is an anode (positively charged) that is \[{{H}^{+}}\] and another is a cathode (negatively charged) that is \[C{{l}^{-}}\].
On the other hand, as sodium chloride which is a weak electrolyte, gets partially ionised as \[N{{a}^{+}}\] and \[C{{l}^{-}}\] and is already present in the solution so, as soon as hydrogen chloride gas is passed and HCl gets ionised, chloride ion concentration in the solution will increase and thus, repulsion created between chloride ions. Due to this repulsion (common ion effect), mostly all chlorine forms bonds with sodium ions again present in the solution, and the solution is left with no more ionised form of sodium chloride. In this way, we can purify sodium chloride.
Thus, the correct option is B.
Note: Buffer solution is that solution whose pH remains constant even if we add even small amount of acid and base (do not allow any change) but in this question, we need to change the impure form of sodium chloride to pure form so, that buffer action does not help to purify the substance. Also, hydrolysis of salt gives a new product (NaOH) instead of pure sodium chloride.
Recently Updated Pages
silver wire has diameter 04mm and resistivity 16 times class 12 physics JEE_Main
A parallel plate capacitor has a capacitance C When class 12 physics JEE_Main
A series combination of n1 capacitors each of value class 12 physics JEE_Main
When propyne is treated with aqueous H2SO4 in presence class 12 chemistry JEE_Main
Which of the following is not true in the case of motion class 12 physics JEE_Main
The length of a potentiometer wire is 10m The distance class 12 physics JEE_MAIN