Honors Practice Test Question
333 mL of 0.444 M sodium hydroxide is mixed with 777 mL of 0.222 M magnesium sulfate.
Find the amount and state (aqueous, liquid, or solid) of each of the product. Determine the
limiting reactant. Determine the final concentrations (that means molarity!) of all of the ionic
species.
58.3 g/mol
2 NaOH (aq) + MgSO4 (aq)  Mg(OH)2 (s) + Na2SO4 (aq)
0.444 M
333 mL
0.222 M
777 mL
xg
Since the amounts of the products are requested, start by finding the limiting reactant and the
grams of the solid first.
58.3 g Mg(OH)2
1 mol Mg(OH)2 0.444 mol NaOH
1 mol Mg(OH)2
2 mol NaOH
58.3 g Mg(OH)2
1 mol Mg(OH)2 0.222 mol MgSO4
1 mol Mg(OH)2
1 mol MgSO4
0.333 L NaOH
= 4.31g Mg(OH)2
1 L NaOH
0.777 L MgSO4 = 10.1 g Mg(OH)2
1 L MgSO4
So, the product that is a solid (the precipitate) is Mg(OH)2 and 4.31 grams of it are formed. The
limiting reactant is NaOH. This means that all of the NaOH is used up in the reaction. The final
request was to calculate the molarity of all the ions left in the solution. So, at the END of the
reaction…..
2 NaOH (aq) + MgSO4 (aq)  Mg(OH)2 (s) + Na2SO4 (aq)
GONE!!
some left over
solid – not ions
made some
So, the following things need to be calculated in order to figure out the molarities:
*how much MgSO4 was at the beginning
*how much MgSO4 was used
*how much MgSO4 was left over
*how much Na2SO4 was made
*the new volume of the solution
Calculate:
*how much MgSO4 was at the beginning:
moles = MV
moles = (0.222 M) (0.777 L)
moles = 0.172 moles
*how much MgSO4 was used:
1 mol MgSO4
0.444 mol NaOH
2 mol NaOH
1 L NaOH
0.333 L NaOH
= 0.0739 mol MgSO4
*how much MgSO4 was left over:
amount left over = initial amount  amount used
amount left over = 0.172 moles  0.0739 moles
amount left over = 0.0981 moles
*how much Na2SO4 was made:
1 mol Na2SO4
0.444 mol NaOH
2 mol NaOH
1 L NaOH
*the new volume of the solution:
0.333 L NaOH
= 0.0739 mol NaSO4
333 mL + 777 mL = 1110 mL= 1.11 L
2 NaOH (aq) + MgSO4 (aq)  Mg(OH)2 (s) + Na2SO4 (aq)
GONE!!
0.0981 moles
1.11 L
solid – not ions
0.0739 moles
1.11 L
Calculate the new molarities and then dissociate them and calculate the molarities of the ions.
MgSO4
Na2SO4
M = moles/volume
M = 0.0981 moles/1.11 L
M = 0.0884 M
M = moles/volume
M = 0.0739 moles/1.11 L
M = 0.0666 M
Write the dissociation reactions:
MgSO4 (aq)  Mg+2 (aq) + SO42 (aq)
0.0884 M
0.0884 M
0.0884 M
Na2SO4 (aq)  2 Na+1 (aq) + SO42 (aq)
0.0666 M
2(0.0666 M)
The final concentrations of the ions are:
Mg+2 = 0.0884 M
Na+1 = 0.133 M
SO42 = 0.0884 M + 0.0666 M = 0.155 M
0.0666 M