How can i calculate the concentration of each ions remaining in the solution after precipitation is complete?
ok heres the full question::
2KOH + Mg(NO3)2 -------------> 2KNO3 + Mg(OH)2
a 100.0 mL aliquot of 0.200 M aqueous potassium hydroxide is mixed with 100.0mL of 0.200M (aq) Magnesium nitrate. it had previously asked for the mass of precipitation produced..and that was 0.583 g Mg(OH)2 . i really don't know how to calculate the concentration of the remaining ions
K+
Mg2+
OH-
NO3-
i know OH- is equal to 0 because its the limiting reagent but for the life of me i cant figure out how to do the rest. could some one please explain it to me.
To calculate the concentration of each ion remaining in the solution after precipitation is complete, you need to consider the stoichiometry of the balanced chemical equation. In this case, the balanced equation is:
2KOH + Mg(NO3)2 → 2KNO3 + Mg(OH)2
Based on the equation, you can determine the mole ratio between the reactants and products. It tells you that for every 2 moles of KOH, you form 1 mole of Mg(OH)2.
First, let's calculate the number of moles of KOH and Mg(NO3)2 used in the reaction:
Moles of KOH = volume (in L) × concentration (in M)
= 0.100 L × 0.200 M
= 0.020 mol
Moles of Mg(NO3)2 = volume (in L) × concentration (in M)
= 0.100 L × 0.200 M
= 0.020 mol
Since the mole ratio between KOH and Mg(OH)2 is 2:1, we can determine the moles of Mg(OH)2 formed as follows:
Moles of Mg(OH)2 = 0.583 g / molar mass of Mg(OH)2
Given the molar mass of Mg(OH)2 is 58.33 g/mol, we can calculate:
Moles of Mg(OH)2 = 0.583 g / 58.33 g/mol
= 0.010 mol
Since the mole ratio between Mg(OH)2 and KOH is 1:2, we can calculate the number of moles of KOH and Mg(NO3)2 remaining after the reaction:
Moles of KOH remaining = Moles of KOH initially - 2 × Moles of Mg(OH)2
= 0.020 mol - 2 × 0.010 mol
= 0.000 mol
Moles of Mg(NO3)2 remaining = Moles of Mg(NO3)2 initially - 1 × Moles of Mg(OH)2
= 0.020 mol - 1 × 0.010 mol
= 0.010 mol
Finally, to calculate the concentration of each ion remaining, divide the moles by the total volume of the solution:
Concentration of K+ remaining = Moles of KOH remaining / Total volume of solution
= 0.000 mol / 0.200 L
= 0.00 M
Concentration of Mg2+ remaining = Moles of Mg(NO3)2 remaining / Total volume of solution
= 0.010 mol / 0.200 L
= 0.050 M
Concentration of NO3- remaining = Moles of Mg(NO3)2 remaining / Total volume of solution
= 0.010 mol / 0.200 L
= 0.050 M
Therefore, after the precipitation is complete, the concentration of K+ remaining is 0.00 M, the concentration of Mg2+ remaining is 0.050 M, and the concentration of NO3- remaining is 0.050 M.