Solution Stoichiometry

Solution stochiometry problems are the same as regular stoichiometry problems except solutions are used. Since solutions are used moles must be determined using molarity and volume.

e.g.

How many grams of NaOH are require to neutralize 37.0 mL of a 0.500 M H2SO4 solution?

To relate an amount of NaOH to an amount of H2SO4 a balanced equation must be used.

Determine the reaction.
Balance the equation.

Since the only way to relate these two chemicals is by using the balanced equation, we must convert everything to moles.

So, we need to find how much NaOH is required to react with 37.0 mL H2SO4. which means we have to relate amount of NaOH to amount of H2SO4. 

So, how many moles of H2SO4 are being used?

Just use molarity as a conversion factor
to convert from mL H2SO4 soln to mol H2SO4.

How many moles of NaOH reaction with that much H2SO4?

Use the balanced equation as a conversion factor
to convert from mol of H2SO4 to mol of NaOH.

And this would be how many grams?

Use molar mass as a conversion factor
to convert from g NaOH to mol NaOH.

= 1.48 g NaOH are needed to neutralize the acid.


Another neutralization reaction

How many mL of a 0.100 M KOH solution are needed to consume 65 mL of a 0.00100 M NaH2PO4 solution?

 

To relate KOH to NaH2PO4 a balanced equation must be used.

Determine the reaction.


The reaction is an acid-base neutralization reaction.
The salt that forms is K2NaPO4.
Balance the equation.

How much NaH2PO4 needs to be consumed?

Start with the number which is measurement,

and use molarity as a conversion.

How much KOH will be needed?

Use the balanced equation to create a conversion factor.

So, how many mL of KOH; afterall, the KOH is coming from a solution...a 0.100 M KOH soln?

Use molarity to convert from mol KOH to mL KOH.


= 1.3 mL of a 0.100 M KOH solution
are required to neutralize the NaH2PO4

Titration problems are another example of stoichiometry problems.

 

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