Transition metals (also called heavy metals) form insoluble compounds with hydroxide ions. Therefore, it is possible to extract heavy metal ions from water by raising the pH. The precipitation reaction between cadmium(II) and hydroxide is shown below.
Cd2+(aq) + 2OH−(aq) → Cd(OH)2(s)
If you had a 0.250L solution containing 0.0130M of Cd2+(aq), and you wished to add enough 1.29M NaOH(aq) to precipitate all of the metal, what is the minimum amount of the NaOH(aq) solution you would need to add? Assume that the NaOH(aq) solution is the only source of OH−(aq) for the precipitation.
How can I set this up? I have part of it set up but I'm not coming out right.
To set up the problem correctly, we need to use the principles of stoichiometry to determine the minimum amount of NaOH(aq) needed to react with all the Cd2+(aq) present in the solution.
First, let's write the balanced equation for the reaction between Cd2+(aq) and OH−(aq):
Cd2+(aq) + 2OH−(aq) → Cd(OH)2(s)
Based on the balanced equation, we can see that 1 mole of Cd2+(aq) reacts with 2 moles of OH−(aq) to form 1 mole of Cd(OH)2(s).
Given the concentration of Cd2+(aq) in the solution, we can calculate the number of moles of Cd2+(aq) using the following equation:
moles of Cd2+ = concentration of Cd2+ × volume of solution
moles of Cd2+ = 0.0130 M × 0.250 L
Now, since the stoichiometric ratio between Cd2+ and OH− is 1:2, we know that we need twice as many moles of OH− as the moles of Cd2+ to completely precipitate all the Cd2+ ions.
moles of OH− needed = 2 × moles of Cd2+
Next, let's calculate the amount of NaOH(aq) solution needed to provide the required moles of OH−. We can use the following equation to calculate the moles of NaOH needed:
moles of NaOH = moles of OH− needed
Finally, we can determine the volume of the 1.29 M NaOH(aq) solution required by dividing the moles of NaOH by its concentration:
volume of NaOH solution = moles of NaOH / concentration of NaOH
Now, let's plug in the values and do the calculations:
moles of Cd2+ = 0.0130 M × 0.250 L = 0.00325 moles
moles of OH− needed = 2 × 0.00325 moles = 0.00650 moles
moles of NaOH = 0.00650 moles
volume of NaOH solution = 0.00650 moles / 1.29 M
Now, divide the moles of NaOH by its concentration to get the volume in liters:
volume of NaOH solution = 0.00650 moles / 1.29 M = 0.0050 L
Therefore, you would need to add a minimum of 0.0050 liters or 5.0 mL of the 1.29 M NaOH(aq) solution to completely precipitate all the cadmium(II) ions in the 0.250 L solution.