what is the molar concentration of the silver ion in the final solution

Question

27.8 mg of silver nitrate (AgNO3, 169.9g/mol) was dissolved in water to make 15.00 mL of solution. 8.15 mL of that solution was diluted to make a new solution with a total volume of 1.000L.

Answer 1

First, find the concentration of AgNO₃ in the original solution:

27.8 mg AgNO₃ x 1 g/1000 mg x 1 mol/169.9 g = 1.636x10^-4 moles AgNO₃

1.636x10^-4 moles AgNO₃ /0.01500 L = 1.09x10^-2 mol/L = 0.0109 M

Now, we can apply our dilution formula: V1M1 = V2M2

(8.15 ml)(0.0109 M) = (1000 ml)(M2) note to change the 1.000 L of solution to 1000 ml to keep units constant

M2 = (8.15 ml)(0.0109 M)/1000 ml

M2 = 0.0000889 M = 8.89x10^-5 M final concentration of AgNO3

Since AgNO₃ dissociates into Ag⁺ and NO₃^-, the concentration of Ag+ ion will be 8.89x10^-5 M