What is concentration of Ag+ ion in a 1 L solution containing 0.02 mol of agNO3 and 0.14 mol of NH3. [Ag(NH3)2]+, Kinstab = 10^8 Options are A. 2 x 10^-7 (M) B. 10^-8 (M) C. 2 x 10^-8 (M) D. 2 x 10^-9 (M)

Dear student,

Second part of your question seems to be incomplete i can't understand what is kinstab and number of moles of which compound is 0.14? So kindly provide complete information and repost your question. Although, the first part of your question is solved below:

Calculation of concentration of Ag⁺:

Number of moles of AgNO_​3 = 0.02 moles
Volume of solution = 1 litres
$Molarity=\frac{moles of solute\left(AgN{O}_3\right)}{Volume of solution}$
$$\begin{gathered} Molarity=\frac{0.02 moles}{1 litre} \\ Molarity=0.02 mol./litre or M \end{gathered}$$
Concentration/molarity of AgNO₃ = 0.02M
Now, as we know that in solution AgNO₃ dissociates into one ion of Ag⁺ and one ion of NO^-₃ .
So, 1 M of AgNO₃ will have 1 M of Ag⁺.
Hence, 0.02 M of AgNO₃ will have 0.02 M of Ag⁺.
Therefore, concentration of Ag⁺ will be 0.02 M.
  Regards