Question #ab2c3

1 Answer
Oct 8, 2015


#"0.25 normal"#


You know that normality is defined as the number of equivalents per liters of solution.

In the case of acids and bases, the equivalents will depend on how many protons, #"H"^(+)#, or hydroxide ions, #"OH"^(-)#, a respective acid or bases releases in solution.

Start by calculating the number of moles of sodium hydroxide by using its molar mass

#50color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g")))) = "1.25 moles NaOH"#

Now, how many moles of equivalents, which for a base are the number of hydroxide ions it releases in solution, do you get per mole of sodium hydroxide?

Well, sodium hydroxide is a strong base, which means that it dissociates completely to produce

#"NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH"_text((aq])^(-)#

For every mole of sodium hydroxide, you produce 1 mole of hydroxide ions. This means that you have

#1.25color(red)(cancel(color(black)("moles NaOH"))) * "1 equivalent"/(1color(red)(cancel(color(black)("mole NaOH")))) = "1.25 equiv. "#

This means that the normality of the solution will be

#N = "no. of equivalents"/"liters of solution"#

#N = "1.25 equiv."/("5 L") = "0.25 normal"#

Notice that you can find a cool relationship between molarity and normality by using the number of equivalents you ger per mole of solute.

In this case, the molarity of the solution would be

#C = n/V#

#C = "1.25 moles"/"5 L" = "0.25 molar"#

When the number of equivalents is equal to thenumber of moles of solute, you have

#N = 1 xx C#

What if you had a solution of #"Ba"("OH")_2#, barium hydroxide?

#"Ba"("OH")_text(2(s]) -> "Ba"_text((aq])^(2+) + 2"OH"_text((aq])^(-)#

The number of equivalents would have been

#1.25color(red)(cancel(color(black)("moles Ba"("OH"_2)))) * "2 equiv."/(1color(red)(cancel(color(black)("mole Ba"("OH"_2))))) = "2.50 equiv."#

The normality would now be twice the molairty

#N = 2 xx C#

This goes to show that you can go from molarity to normality by finding the number of active units, i.e. equivalents, for a particular compound.