Respuesta :
Answer : The precipitate is, AgCl.
Explanation :
The balanced chemical reaction will be:
[tex]AgNO_3(aq)+KCl(aq)\rightarrow AgCl(s)+KNO_3(aq)[/tex]
First we have to calculate the molarity of [tex]Cl^-[/tex] ion.
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the initial molarity and volume of KCl.
[tex]M_2\text{ and }V_2[/tex] are the final molarity and volume of [tex]Cl^-[/tex] ion.
We are given:
[tex]M_1=0.0055M\\V_1=175.0mL\\M_2=?\\V_2=175.0+145.0=320mL[/tex]
Putting values in above equation, we get:
[tex]0.0055M\times 175.0mL=M_2\times 320mL\\\\M_2=0.00301M[/tex]
Now we have to calculate the molarity of [tex]Ag^+[/tex] ion.
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the initial molarity and volume of [tex]AgNO_3[/tex].
[tex]M_2\text{ and }V_2[/tex] are the final molarity and volume of [tex]Ag^+[/tex] ion.
We are given:
[tex]M_1=0.0015M\\V_1=145.0mL\\M_2=?\\V_2=175.0+145.0=320mL[/tex]
Putting values in above equation, we get:
[tex]0.0015M\times 145.0mL=M_2\times 320mL\\\\M_2=0.000679M[/tex]
Now we have to calculate the value of reaction quotient.
The expression for reaction quotient will be :
[tex]Q_{sp}=[Ag^+][Cl^-][/tex]
Now put all the given values in this expression, we get
[tex]Q_{sp}=(0.000679)\times (0.00301)=2.04\times 10^{-6}[/tex]
The given solubility constant value is, [tex]K_{sp}=1.77\times 10^{-10}[/tex]
- When [tex]K_{sp}>Q_{sp}[/tex]; the reaction is product favored. (No precipitation)
- When [tex]K_{sp}<Q_{sp}[/tex]; the reaction is reactant favored. (Precipitation)
- When [tex]K_{sp}=Q_{sp}[/tex]; the reaction is in equilibrium. (Sparingly soluble)
As, [tex]K_{sp}<Q_{sp}[/tex] then the reaction is reactant favored that means formation of precipitation.
Thus, the precipitate is, AgCl.
