Answer :
Answer : The final molarity of iodide anion in the solution is 0.0508 M.
Explanation :
First we have to calculate the moles of [tex]KI[/tex] and [tex]AgNO_3[/tex].
[tex]\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}[/tex]
Molar mass of KI = 166 g/mole
[tex]\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole[/tex]
and,
[tex]\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole[/tex]
Now we have to calculate the limiting and excess reagent.
The given chemical reaction is:
[tex]KI+AgNO_3\rightarrow KNO_3+AgI[/tex]
From the balanced reaction we conclude that
As, 1 mole of KI react with 1 mole of [tex]AgNO_3[/tex]
So, 0.0178 mole of KI react with 0.0178 mole of [tex]AgNO_3[/tex]
From this we conclude that, [tex]AgNO_3[/tex] is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]AgI[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]KI[/tex] react to give 1 mole of [tex]AgI[/tex]
So, 0.0178 moles of [tex]KI[/tex] react to give 0.0178 moles of [tex]AgI[/tex]
Thus,
Moles of AgI = Moles of [tex]I^-[/tex] anion = Moles of [tex]Ag^+[/tex] cation = 0.0178 moles
Now we have to calculate the molarity of iodide anion in the solution.
[tex]\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}[/tex]
[tex]\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M[/tex]
Therefore, the final molarity of iodide anion in the solution is 0.0508 M.