aci bazo

The titrant is the solution of known concentration that is added from the buret... • Step 2: The HCl is placed in the Erlenmeyer flask along with approximately 20.00 mL of distilled wat

Chem 12 Chapter 15 Pg 599-605, 608-611

• Acid-base titration is a process for

calculating the concentration of a known volume of acid or base

ACID-BASE REACTIONS

Titrations

ACID-BASE REACTIONS

Titrations

H2C2O4(aq) + 2 NaOH(aq) ->

acid acid base base

Na2C2O4(aq) + 2 H2O(liq) Carry out this reaction using a

Setup for titrating an acid with a base

Sample Problem

• In this sample titration, we are trying to

determine the concentration of 20.00 mL

of HCl In the titration we will be

neutralizing the HCl with 0.150 M NaOH

• Step 1: The NaOH, the titrant, is placed in

the buret The titrant is the solution of

known concentration that is added from the buret

• Step 2: The HCl is placed in the

Erlenmeyer flask along with approximately 20.00 mL of distilled water and 2-3 drops of phenolphthalein indicator Since the

solution in the flask is acidic,

phenolphthalein is colourless

• Step 3: NaOH is added to the HCl in the

flask When the NaOH comes in contact

with the solution in the flask, it turns pink

and then the pink colour quickly

disappears This is because the OH- from the NaOH interact with the phenolphthalein

to change the phenolphthalein from

colourless to pink

• The solution becomes clear again as the

hydronium ions from the hydrochloric acid neutralize the added hydroxide ions As

more NaOH is added, it takes longer for the pink colour to disappear

• As it starts taking longer for the pink colour

to disappear, the sodium hydroxide is

added a drop at a time

Acid-Base Titration

• The equivalence point of the titration is

reached when equal numbers of moles of hydronium and hydroxide ions have been reacted

• When this happens in this titration, the pH

of the solution in the flask is 7.0 and the

phenolphthalein indicator is colourless

• This would be a good time to stop, however the indicator is still colourless, so must

Titration Curve

• Step 4: Add as little excess NaOH as

possible We want to add a single drop of NaOH to the colourless solution in the flask and have the solution in the flask turn pink and stay pink while the contents of the flask are swirled

• This permanent colour change in the

indicator is known as the endpoint of the

titration and the titration is over

Titration Curve

Solve the problem

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

NaOH mol = 0.150 mol/L x 0.02567 L= 3.85 x 10-3 mol NaOH

Found in titration experiment

concentration of HCl , knowing it is a 1:1 mole ratio

3.85 x 10-3 mol= 0.192 M

SAMPLE PROBLEM 2

• In an acid-base titration, 17.45 mL of 0.180

M nitric acid, HNO3, were completely

neutralized by 14.76 mL of aluminium

hydroxide, Al(OH)3 Calculate the

concentration of the aluminium hydroxide

SAMPLE ANSWER 2

• The balanced equation for the reaction is:

3HNO3(aq) + Al(OH)3(aq) → Al(NO3)3(aq) + 3H2O(l)

• The number of moles of nitric acid used is:

y mol = 0.180 mol/L x 0.01745 L = 3.14 x 10-3 mol HNO3

• From the stoichiometry of the reaction, the number of moles of aluminium hydroxide reacted is:

3.14 x 10-3 mol HNO3 x 1 mol Al(OH)3 = 1.05 x 10-3 mol

3 mol HNO3

• Therefore, the concentration of the aluminium hydroxide is:

1.05 x 10-3 mol Al(OH)3 = 0.0711 M

0.01476 L

• Page 602 # 17-19