|Up until now we have used pure substances and added them to water
in order to make solutions. The point of using solutions is that they
are often easier to use in the lab than pure substances. We will now
find out how to do stoichiometry with these solutions.
|Problem: What volume of 0.556 M HCl
has enough hydrochloric acid to combine exactly with 24.5 mL of aqueous sodium
hydroxide with a concentration of 0.458 M? The equation for the reaction
HCl(aq) + NaOH(aq) -----> NaCl(aq) +
|Solution: This is a simple mL --> mol --> mol --->
Start with what you know: You have 24.5 mL of 0.458 M NaOH. How many moles is this?
0.458 mol = x
x = 0.011 mol of NaOH are being used.
|Step #2 The HCl and NaOH react on a 1:1 basis.
1 HCl = 1 NaOH
x = 0.011 mol of HCl are needed.
|Step #3 We need 0.011 moles of HCl and the solution we
have available is 0.556 M. Therefore
0.556 mol = 0.011 mol
x = 19.78 mL are needed.
If we measure out 19.78 mL of the HCl solution it will
exactly neutralize the NaOH.
|Example Problem #2: How
many millilitres of 0.114 M H2SO4 solution provide
the sulphuric acid required to react with the sodium hydroxide in 32.2 mL
of 0.122 M NaOH according to the following equation?
H2SO4(aq) + 2 NaOH(aq) ------>
Na2SO4(aq) + 2 H2O
|Solution: Again we need the mL --> mol --> mol -->
Start with what you know. You have 32.3 mL of 0.122 NaOH.
0.122 mol = x
x = 0.004 mol of NaOH are present.
1 H2SO4 = 2 NaOH
x 0.004 mol
x = 0.002 mol of H2SO4 are required.
| Step #3 Therefore we need:
0.114 mol = 0.002 mol
Therefore we need 17.54 mL of 0.114 M H2SO4
to exactly neutralize 32.3 mL of a 0.122 M solution of NaOH.
|1. What volume of 0.337 M KOH provides enough solute to combine
with the sulphuric acid in 18.6 mL of 0.156 M H2SO4?
The reaction is:
2 KOH(aq) + H2SO4(aq) ----> K2SO4(aq)
+ 2 H2O
|2. Hydrochloric acid reacts with sodium carbonate as follows:
2 HCl(aq) + Na2CO3(aq) -------> 2 NaCl(aq) + CO2(g) + H2O
What volume of 0.224 M HCl is neutralized
by the Na2CO3 in 24.2 mL of 0.284 M Na2CO3?