In the first semester of general chemistry, we discussed how to balance chemical equations so that we can use the reaction ratio between the species in stiochiometric calculations. Remember that the reaction ratios is the relative number of moles of reactants and products shown in the balanced equation. For example:
HCl + NaOH → NaCl + H2O
HCl only has one acidic hydrogen per formula unit and NaOH only has one hydroxide group per unit, so one formula unit of acid will react with one formula unit of base. We therefore have a 1:1 mole ratio of acid to base. In other cases, for instance the reactions of H2SO4 with NaOH, the mole ration will be 1:2 acid to base (or in other words one unit of acid will react with 2 units of base).
We can use the stiochiometric relationships described here for various stoichiometric calculations. One very important type of calculation involves the reaction of an acid with a base. To represent this reaction and the stiochiometric relationships involved, we will utilize a table method to solve these types of problems.
Molarity Tables Example 1:
100. mL of 0.100 M HCl solution and 100. mL of NaOH are mixed, what is the molarity of the salt in the resulting solution? Assume that the volumes are additive.
To solve this problem, we must first write the balanced equation for the acid-base reaction.
HCl + NaOH → NaCl + H2O
Next we will construct a table that represents Rxn ratio, the number of moles we began with (before), the change in moles (change), the number of moles after the reaction (after), and the new concentration.
Table 1
| HCl | NaOH | NaCl | H2O | |
| Rxn ratio | 1 mol | 1 mol | 1 mol | 1 mol |
| Before | ||||
| Change | ||||
| After | ||||
| New Concentration |
In Table 1 we have equal numbers of moles of HCl and NaOH, therefore all of the HCl and NaOH react. The amount of water produced by the reaction is negligible and is therefore not included in the table or calculations. Next we fill in our given information (Table 2). The change should represent whether the species of interest (reactant or product) is increasing (+) or decreasing (-) and should take the mole ratio into consideration.
Table 2
| HCl | NaOH | NaCl | H2O | |
| Rxn ratio | 1 mol | 1 mol | 1 mol | 1 mol |
| Before | 0.100 L (0.100 M)= 0.0100 mol HCl |
0.100 L (0.100 M)= 0.0100 mol NaOH |
0 mol | |
| Change | -1(0.0100) | -1(0.0100) | +1(0.0100) | |
| After | ||||
| New Concentration |
Finally we complete our calculations to determine the number of moles after the reaction and the new concentration of each reactant and product. Make sure you are adding all of the volumes together to calculation the new concentration.
Table 3
| HCl | NaOH | NaCl | H2O | |
| Rxn ratio | 1 mol | 1 mol | 1 mol | 1 mol |
| Before | 0.100 L (0.100 M)= 0.0100 mol HCl |
0.100 L (0.100 M)= 0.0100 mol NaOH |
0 mol | |
| Change | -1(0.0100) | -1(0.0100) | +1(0.0100) | |
| After | 0 mol | 0 mol | 0.0100 mol | |
| New Concentration | 0 M | 0 M | 0.0100 mol/0.200 L = 0.0500 NaCl |
These calculations can also be expressed in milliliters and millimoles (mmol) rather than in Liters and moles.
Molarity = number of millimoles of solute/ number of milliliters of solution
In the above example the acid and base reacted completely with one another so that only the salt remained in solution. This is not always necessarily the case as is shown in the following example.
Molarity Tables Example 2:
If 100. mL of 1.00 M HBr and 100. mL of 0.80 M KOH solutions are mixed, what are the molarities of the solutes in the resulting solution?
Solution: First write the balanced equation.
HBr + KOH → KBr + H2O
Next enter your givens into the mole table.
Table 4
| HBr | KOH | KBr | H2O | |
| Rxn ratio | 1 mmol | 1 mmol | 1 mmol | 1 mmol |
| Before | 100 mL (1.00 M)= 100. mmol HBr |
100 mL (0.80 M)= 80. mmol KOH |
0 mmol | |
| Change | -1(80. mmol) | -1(80. mmol) | +1(80.00 mmol) | |
| After | 20. mmol | 0 mmol | 80. mmol | |
| New Concentration | 20. mmol/200 mL= 0.10 M HBr |
0 M | 80 mmol/200 mL = 0.40M KBr |