Stoichiometry of Reactions in Solution  
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 is:
 
                     HCl(aq)  +  NaOH(aq) ----->  NaCl(aq)  +   H2O
 
Solution:  This is a simple mL --> mol --> mol ---> mL solution.
 
Step #1
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.
              1000 mL       24.5 mL
 

Step #2    The HCl and NaOH react on a 1:1 basis.  Therefore

                          1 HCl   =     1 NaOH
                              x            0.011 mol

                               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
                           1000 mL              x

                            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 --> mL solution.
 
Step #1
    Start with what you know.    You have 32.3 mL of 0.122 NaOH.

                             0.122 mol  =       x
                               1000 mL       32.3 mL

                               x = 0.004 mol of NaOH are present.
 

Step #2               1 H2SO42 NaOH
                                  x            0.004 mol

                                 x = 0.002 mol of H2SO4 are required.
 

 Step #3     Therefore we need:

                        0.114 mol  =  0.002  mol
                         1000 mL             x                       x = 17.54 mL are needed.

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.
 

Practise Problems
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?
 
 
 
 
 
 
 
 
 

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