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 use the stoichiometry of the balanced chemical equation and the initial concentrations of the reactants.
First, let's write the balanced chemical equation for the reaction:
2KOH + Mg(NO3)2 -> 2KNO3 + Mg(OH)2
From the balanced equation, you can determine the stoichiometric ratio between the reactants and products. This ratio tells you how many moles of each substance are involved in the reaction.
In this case, for every 2 moles of KOH, you get 1 mole of Mg(OH)2. Similarly, for every 1 mole of Mg(NO3)2, you get 1 mole of Mg(OH)2.
Given that you initially mixed equal volumes (100.0 mL) of 0.200 M KOH and 0.200 M Mg(NO3)2, let's calculate the moles of each substance:
For KOH:
moles of KOH = volume of KOH (L) * concentration of KOH (mol/L)
= 0.100 L * 0.200 mol/L
= 0.020 mol
For Mg(NO3)2:
moles of Mg(NO3)2 = volume of Mg(NO3)2 (L) * concentration of Mg(NO3)2 (mol/L)
= 0.100 L * 0.200 mol/L
= 0.020 mol
From the stoichiometry of the balanced equation, we know that for every 2 moles of KOH, we get 1 mole of Mg(OH)2. Since we have 0.020 mol of KOH, we will have 0.010 mol of Mg(OH)2.
Now, let's calculate the concentration of each ion:
K+: Since we started with 0.020 moles of KOH and there is no consumption of K+ ions in the reaction, the concentration of K+ ions after precipitation is complete will be the same as the initial concentration. Therefore, the concentration of K+ ions remaining is 0.200 M.
Mg2+: From the stoichiometry, we know that for every 1 mole of Mg(NO3)2, we get 1 mole of Mg(OH)2. We started with 0.020 moles of Mg(NO3)2, and we know that 1 mole of Mg(OH)2 is produced for every 1 mole of Mg(NO3)2. Therefore, the concentration of Mg2+ ions is also 0.200 M.
OH-: As you correctly stated, OH- ions are fully consumed by the reaction to form Mg(OH)2. So the concentration of OH- ions remaining is zero.
NO3-: Since we started with 0.020 moles of Mg(NO3)2 and there is no consumption or production of NO3- ions in the reaction, the concentration of NO3- ions remaining is the same as the initial concentration. Therefore, the concentration of NO3- ions remaining is 0.200 M.
To summarize, the concentrations of ions remaining in the solution after precipitation is complete are as follows:
K+: 0.200 M
Mg2+: 0.200 M
OH-: 0 M
NO3-: 0.200 M