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 understand the stoichiometry of the reaction and use a series of calculations.
In the balanced equation:
2KOH + Mg(NO3)2 -> 2KNO3 + Mg(OH)2
1. Determine the moles of the reactants used:
Since the initial volume of both solutions is 100.0 mL (0.1000 L) and the concentration is 0.200 M, you can calculate the moles for each reactant:
Moles of KOH = concentration (mol/L) * volume (L) = 0.200 mol/L * 0.100 L = 0.020 moles
Moles of Mg(NO3)2 = concentration (mol/L) * volume (L) = 0.200 mol/L * 0.100 L = 0.020 moles
2. Determine the limiting reagent:
Compare the stoichiometric coefficients of the reactants to determine which one limits the reaction. In this case, the stoichiometric ratio is 2:1 for KOH:Mg(NO3)2.
Since the moles of KOH and Mg(NO3)2 are both 0.020 moles, the ratio is 1:1. Therefore, KOH is the limiting reagent.
3. Calculate the moles of Mg(OH)2 formed:
Since the stoichiometric ratio between KOH and Mg(OH)2 is 2:1, the moles of Mg(OH)2 formed would be half of the moles of KOH used.
Moles of Mg(OH)2 = 0.020 moles / 2 = 0.010 moles
4. Determine the moles of each ion remaining:
Referencing the balanced formula, you can determine the moles of each ion remaining:
Moles of K+ = 0.020 moles * 2 = 0.040 moles
Moles of Mg2+ = 0.020 moles
Moles of NO3- = 0.020 moles
Moles of OH- = 0
5. Calculate the concentration of each remaining ion:
To calculate the concentration, divide the moles of each ion by the total volume of the solution, which in this case is 200.0 mL (0.2000 L):
Concentration of K+ = moles / volume = 0.040 mol / 0.200 L = 0.200 M
Concentration of Mg2+ = moles / volume = 0.020 mol / 0.200 L = 0.100 M
Concentration of NO3- = moles / volume = 0.020 mol / 0.200 L = 0.100 M
Concentration of OH- = 0 M (as it reacted completely with Mg to form the precipitate)
Therefore, the concentration of each ion remaining in the solution after precipitation is complete is:
K+ = 0.200 M
Mg2+ = 0.100 M
NO3- = 0.100 M
OH- = 0 M