Using net ionic equations and solution stoichiometry, calculate the molarity of the principal substances (ions and/or molecules) in solution, or the number of millimoles of insoluble species (solids or gases) present after the following are mixed: YOU MUST USE A TABLE FORMAT AS SHOWN ON THE EXAMPLE PROBLEMS. ) 100.mL of 0.200 M NH4OH and 100. mL of 0.200 M H2SO4
EXAMPLES
Using net ionic equations and solution stoichiometry, calculate the molarity of the principal substances (ions and/or molecules) in solution, or the number of millimoles of insoluble species (solids or gases) present after the following are mixed:
- 200.mL of 0.100 M NaOH and 150. mL of 0.200 M HCl.
Method:
a.) Calculate the millimoles of each ionic or molecular species initially present before mixing.
NaOH completely dissociates to give Na+ and OH- ions so:
200.mL soln x 0.100 mol Na+ (& OH-) = 20.0 mmol Na+ & 20.0 mmol OH-
L soln initially present before mixing
HCl is also 100% ionized so:
150.mL soln x 0.200 mol H+ (& Cl-) = 30.0 mmol H+ & 30.0 mmol Cl-
L soln initially present before mixing
b.) Write the net ionic equation(s) for any reaction(s) occurring between the above species when the solutions are mixed: H+ + OH- -------> HOH
c.) Now consider the stoichiometry. The limiting reactant is
OH-. 20.0 mmol of OH- will react with 20.0 mmol of H+: 20.0 mmol OH- x 1 mmol H+ = 20.0 mmol H+(reacted) 1 mmol OH-
Since 30.0 mmol of H+ was present initially, this leaves 10.0 mmol of H+ remaining after the reaction is complete:
30.0 mmol H+(initially present) – 20.0 mmol H+(reacting) = 10.0 mmol H+(remains)
d.) Calculate the molarity of each species present after reaction. Assume the final volume after mixing is 350. mL (200. mL + 150. mL).
10.0 mmol H+/350. mL = 0.0286 mol H+/L = 0.0286 M H+
20.0 mmol Na+/350. mL = 0.0571 mol Na+/L = 0.0571 M Na+
30.0 mmol Cl-/350. mL = 0.0857 mol Cl-/L = 0.0857 M Cl-
A convenient way to solve this kind of problem is by using a table to summarize the information calculated above:
|
|
Na+ |
OH- |
H+ |
Cl- |
H2O |
|
Initial mmol |
20.0 |
20.0 |
30.0 |
30.0 |
N/A |
|
Change mmol |
0.0 |
- 20.0 |
-20.0 |
0.0 |
+20.0 |
|
Final mmol |
20.0 |
0.0 |
10.0 |
30.0 |
N/A |
|
Final Molarity |
0.0571 |
0.0 |
0.0286 |
0.0857 |
N/A |
2.) 300.mL of 0.150 M KOH and 700. mL of H2O:
Net Ionic Equation: No
|
|
K+ |
OH- |
|
Initial mmol |
45.0 |
45.0 |
|
Change mmol |
0.0 |
0.0 |
|
Final mmol |
45.0 |
45.0 |
|
Final Molarity |
0.0450 |
0.0450 |
3.) 500.mL of 0.250 M HC2H3O2 and 300. mL of 0.100 M Ba(OH)2:
Net Ionic Equation: HC2H3O2 + O H- -------> C2H3O2- + H2O
|
|
HC2H3O2 |
Ba2+ |
OH- |
C2H3O2- |
H2O |
|
Initial mmol |
125 |
30.0 |
60.0 |
0 |
N/A |
|
Change mmol |
-60.0 |
0.0 |
-60.0 |
+60.0 |
+60.0 |
|
Final mmol |
65 |
30.0 |
0.0 |
60.0 |
N/A |
|
Final Molarity |
0.081 |
0.0375 |
0.0 |
0.0750 |
N/A |
(no molarity is calculated for H2O because it is the solvent)
4.) 50.0 mL of 0.100 M Ba(OH)2 and 75.0 mL of 0.200 M H2SO4:
Net Ionic Equation: Two reactions occur
First: H+ + OH- -------> H2O
Second: Ba2+ + SO42- ------> BaSO4(s)
|
|
Ba2+ |
OH- |
H+ |
SO42- |
BaSO4 |
H2O |
|
Initial mmol |
5.00 |
10.0 |
30.0 |
15.0 |
0 |
N/A |
|
Change mmol |
-5.00 |
-10.0 |
-10.0 |
-5.00 |
+5.00 |
+10.0 |
|
Final mmol |
0.00 |
0.0 |
20.0 |
10.0 |
5.00 |
N/A |
|
Final Molarity |
0.00 |
0.0 |
0.160 |
0.0800 |
N/A |
N/A |
(no molarity is calculated for BaSO4 because it is a solid and therefore not in solution)
Using net ionic equations and solution stoichiometry, calculate the molarity of the principal substances (ions and/or molecules) in solution, or the number of millimoles of insoluble species (solids or gases) present after the following are mixed: YOU MUST USE A TABLE FORMAT AS SHOWN ON THE EXAMPLE PROBLEMS.
- ) 100.mL of 0.200 M NH4OH and 100. mL of 0.200 M H2SO4
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