Question #ab2c3
1 Answer
Explanation:
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,
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")_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.
