Question

If 25.8 mL of an AgNO3 solution is needed to precipitate all Cl- ions in a 1570 mg of KCl (forming AgCl), what is the molarity of the AgNO3nsolution?

Answer 1

M=0.816M

Step-by-step explanation:

Hello,

In this case, we should consider the following reaction:

`AgNO_3+KCl\rightarrow KNO_3+AgCl`

Thus, by knowing the 1:1 molar ratio of silver nitrate and potassium chloride, we can easily compute the moles of silver nitrate precipitating the 1570 mg of potassium chloride considering its molar mass of 74.5513 g/mol:

`n_(AgNO_3)=1570mgKCl*(1gKCl)/(1000mgKCl) *(1molKCl)/(74.5513gKCl)*(1molAgNO_3)/(1molKCl) \\\\n_(AgNO_3)=0.021molAgNO_3`

Then, by using the volume of silver nitrate in liters (0.0258 L), we can directly compute the molarity:

`M=(0.021molAgNO_3)/(0.0258L)\\ \\M=0.816M`

Regards.