
In neutralisation of $KI$ by $AgN{{O}_{3}}$ positive charge is due to absorption of
A. $A{{g}^{+}}$ ions
B. $Ag$
C.${{I}^{-}}$ ions
D. both (b) and (c)
Answer
233.4k+ views
Hint: To answer this question we have to know about absorption. Absorption is a process where a gas, liquid is absorbed by another substance. Two types of absorption are present. One of these is physical absorption and another one is chemical absorption.
Complete Step by Step Solution:
When potassium iodide reacts with silver nitrate a neutralisation reaction takes place. In this reaction potassium iodide reacts with silver nitrate to form silver iodide with potassium nitrate. The above reaction is as follows:
$KI+AgN{{O}_{3}}\to AgI+KN{{O}_{3}}$
Thus here silver iodide is formed which can produce $A{{g}^{+}}$ ion or positively charged sol. Thus the positive charge sol is formed due to the absorption of $A{{g}^{+}}$or silver cation.
Again here silver iodide is formed which can produce ${{I}^{-}}$ ion or negatively charged sol. Thus the negative charge sol is formed due to the absorption of ${{I}^{-}}$or iodide ions.
Thus the correct option is A.
Note: Here we have to care about the amount of silver nitrate used. If we use excess silver nitrate then only positively charged sol is formed but if a small amount of silver nitrate is used then both positively charged and negatively charged sols are formed.
Complete Step by Step Solution:
When potassium iodide reacts with silver nitrate a neutralisation reaction takes place. In this reaction potassium iodide reacts with silver nitrate to form silver iodide with potassium nitrate. The above reaction is as follows:
$KI+AgN{{O}_{3}}\to AgI+KN{{O}_{3}}$
Thus here silver iodide is formed which can produce $A{{g}^{+}}$ ion or positively charged sol. Thus the positive charge sol is formed due to the absorption of $A{{g}^{+}}$or silver cation.
Again here silver iodide is formed which can produce ${{I}^{-}}$ ion or negatively charged sol. Thus the negative charge sol is formed due to the absorption of ${{I}^{-}}$or iodide ions.
Thus the correct option is A.
Note: Here we have to care about the amount of silver nitrate used. If we use excess silver nitrate then only positively charged sol is formed but if a small amount of silver nitrate is used then both positively charged and negatively charged sols are formed.
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