Question

A solution is made by dissolving 54.0 g of silver nitrate in enough water to make 350.0 mL. A 10.00 mL portion of this solution is then diluted to a final volume of 250.0 mL. What is the TOTAL concentration of ions present in the final solution?0.908 M1.82 M0.0122 M0.0726 M0.0363 M

Answer 1

Total concentration of ions present in the final solution is 0.0726 M.

Step-by-step explanation:

Mass of silver nitrate = 54.0 g

Moles of silver nitrate =
`(54.0 g)/(170 g/mol)=0.318 mol`

Volume of the solution made = V = 350.0 mL = 0.350 L

`Molarity=(Moles)/(Volume(L))`

`M=(0.318 mol)/(0.35 L) =0.9086 M`

After dilution of 10.00 mL 0.9227 M solution.

`M_1=0.9086 M`

`V_1=10.00 mL`

`M_2=?`

`V_2=250.0 mL`

`M_1V_1=M_2V_2`

`M_2=(M_1V_1)/(V_2)=(0.9086 M* 10.00 mL)/(250.0 mL)`

`M_2=0.0363 M`

Concentration of silver nitrate after dilution = 0.0363 M

`AgNO_3\rightarrow Ag^++NO_3^(-)`

`[AgNO_3]=[Ag^+]=[NO_3^(-)]=0.0363 M`

Total concentration of ions present in the final solution:

`[Ag^+]+[NO_3^(-)]=0.0363 M+0.0363 M=0.0726 M`