# Question ab2c3

Oct 8, 2015

$\text{0.25 normal}$

#### 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, ${\text{H}}^{+}$, or hydroxide ions, ${\text{OH}}^{-}$, 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

${\text{NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

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

$N = \text{no. of equivalents"/"liters of solution}$

$N = \text{1.25 equiv."/("5 L") = "0.25 normal}$

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

$C = \frac{n}{V}$

$C = \text{1.25 moles"/"5 L" = "0.25 molar}$

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

$N = 1 \times C$

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

${\text{Ba"("OH")_text(2(s]) -> "Ba"_text((aq])^(2+) + 2"OH}}_{\textrm{\left(a q\right]}}^{-}$

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

$N = 2 \times C$

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.