Answer : The final molarity of iodide anion in the solution is 0.0508 M.
Explanation :
First we have to calculate the moles of
and
.
![\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DKI%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DKI%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DKI%7D)
Molar mass of KI = 166 g/mole
![\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DKI%3D%5Cfrac%7B2.95g%7D%7B166g%2Fmole%7D%3D0.0178mole)
and,
![\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DAgNO_3%3D%5Ctext%7BConcentration%20of%20%7DAgNO_3%5Ctimes%20%5Ctext%7BVolume%20of%20solution%7D%3D0.0620M%5Ctimes%200.350L%3D0.0217mole)
Now we have to calculate the limiting and excess reagent.
The given chemical reaction is:
![KI+AgNO_3\rightarrow KNO_3+AgI](https://tex.z-dn.net/?f=KI%2BAgNO_3%5Crightarrow%20KNO_3%2BAgI)
From the balanced reaction we conclude that
As, 1 mole of KI react with 1 mole of ![AgNO_3](https://tex.z-dn.net/?f=AgNO_3)
So, 0.0178 mole of KI react with 0.0178 mole of ![AgNO_3](https://tex.z-dn.net/?f=AgNO_3)
From this we conclude that,
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 ![AgI](https://tex.z-dn.net/?f=AgI)
From the reaction, we conclude that
As, 1 mole of
react to give 1 mole of ![AgI](https://tex.z-dn.net/?f=AgI)
So, 0.0178 moles of
react to give 0.0178 moles of ![AgI](https://tex.z-dn.net/?f=AgI)
Thus,
Moles of AgI = Moles of
anion = Moles of
cation = 0.0178 moles
Now we have to calculate the molarity of iodide anion in the solution.
![\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7DAgNO_3%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DAgNO_3%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D)
![\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7DAgNO_3%3D%5Cfrac%7B0.0178mol%7D%7B0.350L%7D%3D0.0508M)
Therefore, the final molarity of iodide anion in the solution is 0.0508 M.