Honors Chemistry I Titration Lab Calculations Remember that in Chapter 19, Part 2, we learned about acid-base neutralization: Acid + Base Salt + Water e.g. HCl + NaOH NaCl + H2O When you titrate HCl with NaOH, they react with 1:1 stoichiometry (a 1:1 mole ratio) Moles acid = Moles base (Molarity x Volume)acid = (Molarity x Volume)base So, for example, from the data for a titration, we can calculate the molarity of the base if we know the molarity and volume of the acid and the volume of the base. Remember the product of molarity and volume is equal to moles of acid.

Molarity Volume

Volumemoles

VolumeMolarityacid acid

base

acid

basebase

•= =

Sometimes acids have more than one proton to donate and the reaction they undergo depends on how much base you add. For diprotic acids, you must add twice as much base to neutralize both protons. Also, some bases yield more than one mole of OH- ions, like 1 mole of Ca(OH)2 gives 2 moles of OH-, and 1 mole of Al(OH)3 gives 3 moles of OH-. Calculating the amount of acid needed to neutralize a given amount of base is made easier if we define a unit called an equivalent. One equivalent is the amount of acid or base that will give one more of H3O+ or OH- ions. The equivalent mass is the mass of one equivalent of an acid or base. In this titration experiment you will determine the equivalent mass of an unknown acid – that is, the number of grams of the acid that will supply one mole of H3O+ ions. For example: 1 mole HCl = 1 equivalent HCl 1 mole H2SO4 = 2 equivalents H2SO4 1 mole NaOH = 1 equivalent NaOH 1 mole Ca(OH)2 = 2 equivalents Ca(OH)2 In an titration at the equivalence point: Equivalents acid = Equivalents base (but moles of acid might not = moles of base) The only time when moles of acid = moles of base is when the equivalents are equal to one each, such as NaOH and HCl. Since we use equivalents instead of moles, we define concentration unit as Normality: Normality = equivalents of solute / liter of solution (note that Molarity = Normality for a solution of any acid that supplies 1 mole of protons per mole of acid or any base that supplies 1 mole of hydroxide ions per mole of base) So 1.0 M NaOH = 1.0 N NaOH and 1.0 M HCl = 1.0 N HCl But 1.0 M H2SO4 = 2.0 N H2SO4

Calculations Required for Titration Lab:

1. Moles of KHP (Potassium Hydrogen Phthalate – molar mass = 204.22 grams/mole = 204.22 grams/equivalent). Since molar mass and equivalent mass are the same, then this acid is monoprotic; meaning that it only gives up one hydrogen and moles = equivalents.

2. Normality of NaOH in your 2-liter bottle (you determined this by titrating the known KHP with your

NaOH solution):

In any titration at the equivalence point: Equivalents acid = Equivalents base Normalityacid x Volumeacid = Normalitybase x Volumebase

Equivalentsacid = Normalitybase x Volumebase Since we know that KHP is monoprotic Molesacid = Normalitybase x Volumebase

molesvolume

Normalityacid

basebase=

3. Percent Deviation:

Simply take the difference between the consecutive trials divided by the average between the two consecutive trials. trial trial

Avgdeviation

trials

1 2 100−

• = %

You will need to be confident in your results, therefore you will need to achieve less than 1% deviation to be able to move on.

4. Equivalent Mass of your unknown acid:

Equivalents acid = Equivalents base

gramequivalent mass

unknownunkown

acidacid

Normality Lbase base= •

We can rearrange this expression to solve for the equivalent mass of the unknown acid.

Equivalent mass unknown acid (in grams per equivalent) = ( )grams

NormalityunknownLiters

acid

base base•

You will need to do AT LEAST TWO titrations of your unknown to be confident about your results.