Example: Balance the equation above in basic solution (I don't believe this reaction actually occurs in basic solution, but due to our time constraints we'll pretend it does.)
First we balance as above to give:
8 H2O + 2 MnO4- + 10 Cl- 2 Mn2+ + 5 Cl2 + 16 OH-
The most commonly used concentration term in chemistry is molarity = 1 mole of solute dissolved in enough solvent to give 1 L = M = mol/L. This is the most popular concentration unit at least in part because of the convenience of making molar solutions with volumetric flasks (show flask).
Note that molarity has various meanings in describing solutions. Commonly molarity refers to the amount of substance dissolved in one liter of liquid to give a solution. Thus if 158.52 g (1 mole) of strontium chloride is dissolved in enough water to give one liter we will have a 1 L solution of strontium chloride. Note however, that while the solution is 1M in Sr2+, it is 2 M in Cl- and 3 M in terms of particles!
There are a number of different kinds of problems.
Make up a 1.000 L solution of 0.25 M NaCl (note that water is the "default" solvent).
What is the concentration of a solution made by dissolving 10.00 g of KI in enough water to make 1.000 L
What is the concentration of a solution resulting from adding 25.0 mL of 0.60 M CaCl2 to 475 mL of water. (Note - both aqueous, so the volumes are additive.)
First, I like to keep in mind that Moles = Moles, that is we have conservation of mass.
What is the concentration of chloride ion in this solution?
How much 1.000 M MgSO4 is needed to make 500.0 mL of a 0.25 M solution.
When we combine equal numbers of moles of hydrogen ion and hydroxide ion a neutralization occurs. That is, there is no reactive component left, all of the acid has been consumed by all of the base, and water has been synthesized.
Consider the reaction of 50.0 mL of 0.25 M hydrochloric acid with 25.0 mL of 0.50 M sodium hydroxide.
- Write a net ionic equation for this reaction:
H+ + Cl- + Na+ + OH- H2O + Cl- + Na+
Giving: H+ + OH- H2O
Consider the reaction of a weak acid and strong base: 50.0 mL of 0.25 M acetic acid is reacted with 18.0 mL of 0.50 M sodium hydroxide. Find the number of moles of each of the reactants and products after reaction.
- First recall that for the reaction of a weak acid and a strong base the reaction will go until one of the reactants is completely consumed.
- Writing the net ionic reaction we get:
CH3COOH + OH- H2O + CH3COO-
- First find moles of each reactant:
- acid = (50.0 mL)(1 L/1000 mL)(0.25 mole/L) = 1.25 x 10-2moles
- base = (18.0 mL)(1 L/1000 mL)(0.50 mole/L) = 0.900 x 10-2moles
- After reaction all of the base is consumed, so:
- base = 0
- acetate = 0.900 x 10-2moles
- acid = 1.25 x 10-2moles - 0.900 x 10-2moles = 0.35 x 10-2moles
- Water synthesized = 0.900 x 10-2moles.
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© R A Paselk
Last modified 14 February 2011