Determine whether each compound is soluble or insoluble. If the compound is soluble, list the ions present in solution.

a. AgNO3 b. Pb(C2H3O2)2 c. KNO3 d. (NH4)2S

Solution: Here, we are going to calculate the volume of stock solution used to prepare 1000 mL of 0.250 M ammonia. Also, we are going to find the concentration of the final solution if 10 mL of the stock solution is diluted to 0.500 L. Step1: In order to dilute a stock solution, we have to use the following dilution equation, M 1 1M V 2 2 -----------(1) Here, M an1V are t1 molarity and volume of the concentrated stock solution whereas M and 2 V are the molarity and volume of the diluted solution. 2 Step2: a) In the above problem, let us consider the stock solution of NH to be concentrated 3 solution and the solution of NH to be prep3ed as dilute solution. Therefore, M (con1ntrated) = 14.8 M V (concentrated) = 1 M 2ilute) = 0.250 M V 2ilute) = 1000.0 mL Putting these values in equation (1), we get, 14.8 M X V = 0.10 M X 1000 mL 14.8 M X V = 250 (M x mL) 1 V1 250 / 14.8 [(M x mL) / M] = 16.89 mL Thus, 16.89 mL of stock solution is required to make 1000.0 ml of 0.250 M ammonia. Step3: b) Again, for the second problem, let us consider the stock solution of NH to be 3 concentrated solution and the solution of NH to be prepared 3 dilute solution. Therefore, M (concentrated) = 14.8 M 1 V 1oncentrated) = 10.0 mL M 2ilute) = V (dilute) = 0.500 L 2 = 0.500 x 1000 mL = 500 mL Putting these values in equation (1), we get, 14.8 M X 10.0 mL = M X 500 mL 2 148 (M x mL) = M X 500 mL2 M 2148 / 500 [(M x mL) / mL] = 0.296 M Thus, the concentration of the final solution will be 0.296 M. ----------------------