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 apply the principles of stoichiometry and use the information provided in the question.

Let's start by determining the moles of magnesium hydroxide (Mg(OH)2) formed. You previously calculated the mass of the precipitate as 0.583 g, which we can convert to moles using the molar mass of Mg(OH)2.

The molar mass of Mg(OH)2 is calculated as follows:
Mg = 24.31 g/mol
O = 16.00 g/mol (there are two oxygen atoms)
H = 1.01 g/mol (there are two hydrogen atoms)

Molar mass(Mg(OH)2) = Mg + 2(O) + 2(H) = 24.31 + 2(16.00) + 2(1.01) = 58.33 g/mol

Now, let's calculate moles of Mg(OH)2:
moles = mass / molar mass = 0.583 g / 58.33 g/mol = 0.01 mol

Since the balanced equation shows that 2 moles of KOH react with 1 mole of Mg(NO3)2 to produce 1 mole of Mg(OH)2, we can now determine the moles of Mg(NO3)2 reacted:

moles of Mg(NO3)2 = 0.01 mol / 1 mol = 0.01 mol

Since we initially have 100.0 mL of 0.200 M (mol/L) Mg(NO3)2, the total moles of Mg(NO3)2 can be calculated:

moles of Mg(NO3)2 = volume (L) x concentration (mol/L)
= 0.100 L x 0.200 mol/L = 0.02 mol

According to the balanced equation, 1 mole of Mg(NO3)2 produces 1 mole of Mg(2+). Therefore, the concentration of Mg(2+) remaining in the solution will be:

Mg(2+) = (moles of Mg(NO3)2 - moles of Mg(OH)2) / total volume (L)

Mg(2+) = (0.02 mol - 0.01 mol) / 0.200 L
= 0.01 mol / 0.200 L
= 0.05 M

Therefore, the concentration of Mg(2+) remaining in the solution is 0.05 M.

Similarly, the concentration of K+ remaining in the solution can be calculated as follows:

K+ = (moles of KOH - moles of Mg(NO3)2) / total volume (L)

Since 2 moles of KOH react with 1 mole of Mg(NO3)2, the moles of KOH reacted are twice the moles of Mg(NO3)2 reacted. Hence:

moles of KOH = 2 x moles of Mg(NO3)2
= 2 x 0.01 mol = 0.02 mol

K+ = (0.02 mol - 0.01 mol) / 0.200 L
= 0.01 mol / 0.200 L
= 0.05 M

Therefore, the concentration of K+ remaining in the solution is also 0.05 M.

Finally, since OH- is the limiting reagent in this reaction and reacts completely with Mg(2+), the concentration of OH- remaining in the solution is 0 M.

As for NO3-, it is a spectator ion and therefore does not participate in the precipitation or react in any way. Thus, the concentration of NO3- remaining in the solution is the same as the initial concentration, which is 0.200 M.

To summarize, after precipitation is complete, the concentration of each ion remaining in the solution is:
K+ = 0.05 M
Mg(2+) = 0.05 M
OH- = 0 M
NO3- = 0.200 M