Chemistry 11 – Notes on Chemical Reactions

8

Chemistry 11 – Acid / Base Titration Notes

At the beginning of the year we learned how to classify chemical reactions. One of the special double replacement reactions that we came across was called a “Neutralization Reaction.” Remembering back, a neutralization reaction involves a reaction between an acid and a base to give us a salt and water.

Acid + Base ! Salt + Water

In general, acids are compounds that begin with an “H” in their formula. (Organic acids ends with “COOH”.) The following is a list of common acids that you will often see in a chemistry lab:

HF Hydrofluoric Acid H2SO4 Sulphuric Acid

HCl Hydrochloric Acid H2SO3 Sulphurous Acid

HBr Hydrobromic Acid HNO3 Nitric Acid

HI Hydroiodic Acid HNO2 Nitrous Acid

H3PO4 Phosphoric Acid CH3COOH Acetic Acid

H2CO3 Carbonic Acid In general, bases are compounds that end with an “OH” in their formula and are commonly known as hydroxides. The following is a list of common bases that you will often see in a chemistry lab:

NaOH Sodium Hydroxide NH4OH Ammonium Hydroxide

Ca(OH)2 Calcium Hydroxide Al(OH)3 Aluminum Hydroxide

In general, a salt is defined as a substance that is neither an acid nor a base. It is an ionic

compound that does not begin with an “H” or end with an “OH.” (Example: NaCl)

Thus, an example of a neutralization reaction that involves HCl and NaOH is as follows:

HCl + NaOH ! NaCl + HOH Acid Base Salt Water

acid

base salt water

HZO

Acid Base Salt

acid beginCOOH

sulfphate

142804

T

solarsulfite

=

sulphurous Acid

BILL,Ym§

" "

Hydrobromic Acid -nitrate Nitric Acid

← Phosphate

HNOZ

H3P04= Vinegar

= acetic acid

Carbonic Acid

C032'

= Carbonate

base end OH

hydroxide

fmetal

NH4+ OH-

NAOH NH4OH

T.me,ae

Calcium hydroxide A|( OH )3Aptat

salt acid base

doesn't begin H end OH

ACID BASE SALT water

Chemistry 11 – Notes on Chemical Reactions

9

Titrations Chemists have developed a special process of finding an unknown concentration of a

chemical in a solution based upon the neutralization reaction. This process is known as

the “Titration.”

In a titration, a solution of known concentration is reacted with a solution of an

unknown concentration (but a known volume), until a desired equivalence point is

reached.

The equivalence point is the point at which all of the acid and base have reacted together

in a solution. In other words, the moles of acid present is equal to the moles of base

present in the solution!

Moles of Acid = Moles of Base

Thus, all that is left in the solution after the reaction is salt and water.

The following is a diagram of a typical acid / base titration experiment setup:

titration ↳ ACID t BASE → SALT + WATER

← M = MOI

known concentrationL

Unknown concentration equivalence point

moles of base

present

Mole Moll

salt water

@← burette

me← 0.0mL = initial

÷

= final } final - initial

in← 15.0mL 15.0mL - 0.0mL

NAOH =

- =tttrant =

= -115.0mL÷

Known ÷=

[ ] and÷←50.0mL

Volume T*

He , •

← Erlenmeyer flask

analyte 11

unknown -L # Demo

.-

- - --

.. [ NAOH ] = O.IMKnown ✓

-# VNAOH = V final - Vinitial

- = 24.40Mt - 13.62mL

= 10.78mL

[ HCI ] = ? unknown

VHCI = 10.0mL

Chemistry 11 – Notes on Chemical Reactions

10

We use a table and stoichiometry to help us figure out the unknown concentration

after we have collected the data from the experiment.

Example: Consider the reaction: H3PO4 + 2KOH ! K2HPO4 + 2H2O If 0.0198 L of H3PO4 with an unknown concentration is reacted with 0.0250 L of 0.500 M KOH to reach the equivalence point, what is the concentration of H3PO4? Table: H3PO4 KOH

Molarity ? 0.500 M

Volume 0.0198 L 0.0250 L

Road Map: Conc KOH ! Moles KOH ! Moles H3PO4 ! Conc H3PO4 Step 1: Calculate the moles of KOH present. Moles KOH = 0.500 M x 0.0250 L = 0.0125 mol KOH Step 2: Calculate the moles of H3PO4 present. Moles H3PO4 = 0.0125 mol KOH x 1 mol H3PO4 = 0.00625 mol H3PO4 2 mol KOH Step 3: Calculate the concentration of H3PO4. Conc H3PO4 = 0.00625 mol H3PO4 = 0.316 M 0.0198 L In a titration experiment, usually more than one volume data is taken to make sure that the

titration is done accurately. How do we approach a question like this? Let’s take a look at

an example.

Example: Consider the reaction: Ca(OH)2 + 2HCl ! CaCl2 + 2H2O A lab technician titrates a 10.0 mL sample of Ca(OH)2 solution with 0.0156 M HCl. The following are the volume data for the titration:

Trial Volume HCl used 1 15.35 mL 2 15.45 mL 3 15.80 mL

table stoichiometry

ACID BASE SALT

1 2 1 2

-

I know more about KOH

8

am:O"

:B

.la#I@3SF3SF3SF

mol KOH = 0.500 m[lK0H_ ×0.0250L KOH = 0.0125 mol

KOHKOH

as

mol H3P04=

0.0123¥MOIKOH ×

1*+3104=6.2355×40-3

mol KOH mol H3P04

3SF

[H3P04 ] = MOI H3PO4=

6.25×10 -3mg H3PO4- - = 0.3156T MH3PO¢

L H3PO4 0.0198 L H3P04= 0.316N

ZSF

volume

accurately

× = outlier

Chemistry 11 – Notes on Chemical Reactions

11

What is the molarity of the Ca(OH)2 in the solution? Note: If you are given more than one trial for the volume data, choose at least two trials

that have the volume within ± 0.10 mL and take an average of the two volumes.

Average Volume used = (15.35 + 15.45) mL = 15.40 mL HCl 2 Table: HCl Ca(OH)2

Molarity 0.0156 M ?

Volume 15.40 mL 10.0 mL

Road Map: Conc HCl ! Moles HCl ! Moles Ca(OH)2 ! Conc Ca(OH)2 Step 1: Calculate the moles of HCl present. Moles HCl = 0.0156 M x 0.01540 L = 0.0002402 mol Step 2: Calculate the moles of Ca(OH)2 present. Moles Ca(OH)2 = 0.0002402 mol HCl x 1 mol Ca(OH)2 = 0.0001201 mol 2 mol HCl Step 3: Calculate the concentration of Ca(OH)2. Conc Ca(OH)2 = 0.0001201 mol Ca(OH)2 = 0.0120 M 0.0100 L

more than omtrial

10.10mL average

15.35mL + 15.45mL

2= 15.40mL HC/

LHCI + CALOH )z → Caclzt 2h20

f start

Ha Cal OH )z

[ ] 0.0156N ?

Volume# 15.40mL 10.0mL No need to

a- volume units match = ML → L

[ ] mol mol [ ]# mroateio #

mol = M x L = 0.0156 molttclx15.40mL HU

= 0.24024µmolHaL Hel

0.240241mF#× teeth2mot# 0-12012 a mol Cal OH )z

M= mol ÷ L

1

0.12012 _m/mol CALOH )z× -10.0µL CALOH)z

= 0.012012*472= 0.012012 M =

0.0120mLCACOH )z