Write balanced complete ionic and net ionic equations for each reaction.
Solution: Here, we are going to calculate the mass of NaOH needed to precipitate the cadmium ions from 35.0 mL of 0.500 M cadmium nitrate solution. Step1: The reaction between NaOH and Cd(NO ) takes pl3 2n the following manner. 2NaOH(aq) + Cd(NO ) (aq3 2----> Cd(OH) (aq) + 2aNO (aq) 3 In the above reaction, we have seen that 1 mole of Cd(NO ) require3 2oles of NaOH to precipitate out all the cadmium ions. Step2: Now, let us calculate the number of moles of 35.0 ml of 0.500 M Cd(NO ) solution. 3 2 Number of moles of Cd(NO ) solution = Volume of solution in litres X Molarity 3 2 = (35/1000 L) X 0.500 M = 0.035 L X 0.500 M = 0.0175 mol Thus, 0.0175 mol of Cd(NO ) sol3 2 will require (2 x 0.0175 = 0.035) mol of NaOH to precipitate out all the cadmium ions. Step3: Now, 1 mole of NaOH = molar mass of NaOH Therefore, 0.035 mol of NaOH = 0.035 X molar mass of NaOH = 0.035 x 40.01 g [molar mass of NaOH = 40.01 g/mol] = 1.40 g Thus, 1.40 g of NaOH is needed to precipitate the cadmium ions from 35.0 ml of 0.500 M Cd(NO ) solution. 3 2