Answer:
Total concentration of ions present in the final solution is 0.0726 M.
Explanation:
Mass of silver nitrate = 54.0 g
Moles of silver nitrate = [tex]\frac{54.0 g}{170 g/mol}=0.318 mol[/tex]
Volume of the solution made = V = 350.0 mL = 0.350 L
[tex]Molarity=\frac{Moles}{Volume(L)}[/tex]
[tex]M=\frac{0.318 mol}{0.35 L} =0.9086 M[/tex]
After dilution of 10.00 mL 0.9227 M solution.
[tex]M_1=0.9086 M[/tex]
[tex]V_1=10.00 mL[/tex]
[tex]M_2=?[/tex]
[tex]V_2=250.0 mL[/tex]
[tex]M_1V_1=M_2V_2[/tex]
[tex]M_2=\frac{M_1V_1}{V_2}=\frac{0.9086 M\times 10.00 mL}{250.0 mL}[/tex]
[tex]M_2=0.0363 M[/tex]
Concentration of silver nitrate after dilution = 0.0363 M
[tex]AgNO_3\rightarrow Ag^++NO_3^{-}[/tex]
[tex][AgNO_3]=[Ag^+]=[NO_3^{-}]=0.0363 M[/tex]
Total concentration of ions present in the final solution:
[tex][Ag^+]+[NO_3^{-}]=0.0363 M+0.0363 M=0.0726 M[/tex]
