Question:A solution of AgNO3 (45 mL/0.45 M) was mixed with solution of NaCl (85 mL/1.35 x 10-2 M)

Question

a) Calculate the ion product of the potential precipitate.

My answer:

mol AgNO3 = M x L
          = 0.45 M x 0.045 M
          = 0.02025

mol NaCl = M x L
         = 0.0135 M x 0.085 L
         = 0.0011475

(Ag^+) = mols / total volume 
       = 0.02025 / 0.130 L
       = 0.1558

total volume = 85 mL + 45 mL 
             = 130 mL
             = 0.130 L

(Cl^-) = mols / total volume 
       = 0.0011475 / 0.130 L
       = 0.00883

Ion product = [Ag^+][Cl^-]
            = [0.1558][0.00883]
            = 0.001375

Is my answer correct?

Abigail · Answer

b) Would a precipitate form? The Ksp of AgCl(s) is 1.8 x 10-10. Also, please check if my answer to be is also correct or not When comparing the ion product to Ksp, I see that 0.001375 is greater than 1.8 x 10^-10, therefore a precipitate will form.

Explain Bot · Answer

Yes, your answer is correct. You correctly calculated the moles of AgNO3 and NaCl by multiplying the molarity by the volume in liters. Then, you determined the concentration of Ag+ and Cl- ions by dividing the moles by the total volume in liters. Finally, you calculated the ion product by multiplying the concentrations of Ag+ and Cl- ions.