Calculating molarity and normality?

Question

A solution of citric acid (a triprotic acid, C6H8O7) is prepared by dissolving 22.54g in water and made up to 500.0mL.

What is the molarity of this solution?

What is its normality?

When this acid solution is used to titrate a 25.00mL sample of NaOH, it takes 18.55mL acid to reach an end point. What is the molarity of the NaOH solution?

Sorry to bombard you with all these questions, but my chemistry teacher is horrible and he refuses to help me catch up after being out for a week (in the hospital). Any help or explanation you could give me would be wonderful.

Answer 1

For molarity (M) you find how many moles are in 22.54 grams of this substance. To do that, divide 22.54 by the molecular weight of citric acid (I think about 192 grams per mole). Then divide 500.0mL by 1000 to find the number of liters of water. Then divide your number of moles found before by the number of liters just calculated. This will give you molarity of the acidic solution (in moles per liter). Unfortunately, I haven't learned "normality" so I can't help you there.

Answer 2

Molarity is the moles of solute per 1 L of solution
so M=mole/V with V expressed in L
Also mole=mass/MW so you have
M=mass/(MW*V)

The molecular weight (MW) of citric acid is 192 so the Molarity is
M =22.54/(192*0.5) = 0.235

Normality is a very useful way to express concentration, since it represents the concentration of "functional" chemical groups.

What does this mean?

Let's take the case of acids and bases. When an acid and a base react together the net reaction is actually

H+ +OH- ->H2O

while the remaining parts of the acid and base molecules form a salt.
So in acid-base reaction the fundamental units for the reaction are the H+ and OH- ions. Thus it is helpful to express the amount of acid/base in terms of their content in H+/OH- respectively. In order to do this the term (chemical) equivalent is used. 1 equivalent is equal to 1 mole of the fundamental unit for the reaction. So 1 acid equivalent is equal to 1 mole H+.

When you have acids that can give more than 1 H+ per molecule then you realize that 1 mole of the acid contains more than one mole H+: A triprotic acid gives 3 H+, so 1 mole of the acid will contain 3 mole H+ and thus 3 acid equivalents.

Normality (N) is the number of equivalents per 1L of solution
So if the valency (number of H+ given) of the acid is "a", you have the equations

eq= a*mole,
N=a*M
and eq=N*V

In this case (triprotic acid) a=3 so N =3*0.235 = 0.705.

For a base "a" would be the number of OH- per molecule.

In redox reactions the fundamental unit of the reaction is the electron, thus "a" would be the number of electrons gained/lost during the reaction.

The usefulness of Normality is seen in titrations.
At the end-point the equivalents of the acid are equal to the equivalents of base :
eq(acid)= eq(base)=>
N(acid)* V(acid) =N(base)*V(base) (1)

But N=a*M, and NaOH has 1 OH- in its molecular formula thus it has a=1 and N(base)=1*M(base) = M(base).
So equation (1) is re-written

N(acid)* V(acid) =M(base)*V(base) =>
M(base) =N(acid)* V(acid)/ V(base) =0.705*18.55/25 =0.523

Note that I didn't convert mL into L because the convertion factor is simplified in the ratio.
Also equation (1) can be written in the form

a1*M1*V1= a2*M2*V2

and can be used in acid/base or redox titrations accordingly to calculate the molarity of any of the unknowns.