Question

What volume, in ml, of a 0.2089 m ki solution contains enough ki to react exactly with the cu(no3)2 in 43.88 ml of a 0.3842 m solution of cu(no3)2?

Answer 1

The reaction is given as: $2Cu(NO_{3})_{2}+4KI\rightarrow 2CuI+I_{2}+4KNO_{3}$ Here, two moles of copper nitrate reacts with four moles of potassium iodide to give two moles of copper iodide, one mole of iodine and four moles of potassium nitrate. First, calculate the number of moles of copper nitrate. Number of moles is equal to the product of molarity and volume of solution in litre. Number of moles = $0.3842 M\times 0.04388 L$    (1 L =1000 mL)

= $0.016858696 mole$

Copper nitrate requires = $0.016858696 \times \frac{4}{2}$ mole of potassium iodide

= $0.033717392 mole$ of potassium iodide

Volume of solution in litre = $\frac{number of moles}{Molarity}$

Thus, volume of potassium iodide is  =$\frac{0.033717392}{0.2089}$

= $0.1614 L$

1 L =1000 mL

Volume of potassium iodide in mL =$161.4 mL$

Hence, $161.4 mL$ 0.2089 M potassium iodide consist of sufficient potassium iodide to react with copper nitrate in 3.88 mL of a 0.3842 M solution of copper nitrate .