Download Sample Problem Acid-Base Titration 1

• Acid-base titration is
a process for
calculating the concen
tration of a known
volume of acid or base.
H
(
4
0
C
2
aq)
acid
+
2 NaOH(aq)
—>
2
N
(
4
0
aq
aC
)
+
22
8
0
(Iiq)
base
Carry out this reaction
using a TITRATION.
Setup for titrating an
acid with a base
Sample Prob
lem
• In this sample titration
, we
determine the concentra are trying to
HCI. In the titration we wil tion of 20.00 mL of
HCI with 0.150 M NaOH. l be neutralizing the
• Step 1: The NaOH, the
buret. The titrant is the titrant, is placed in the
concentration that is addsolution of known
ed from the buret.
• Step 2: The HCI is plac
ed in the Erlenmeyer
flask along with approxima
distilled water and 2-3 dro tely 20.00 mL of
ps of phenolphthalein
indicator. Since the solution
phenolphthalein is colourle in the flask is acidic,
ss.
• Step 3: NaOH is added to
the HCI in the flask.
When the NaOH comes in
contact with the
solution in the flask, it turn
s pink and then the
pink colour quickly disapp
ears. This is because
the OH- from the NaOH inte
ract with the
phenolphthalein to change
the phenolphthalein
from colourless to pink.
• The solution becomes cle
ar again as the
hydronium ions from the
hydrochloric acid
neutralize the added hydrox
ide ions. As more
NaOH is added, it takes lon
ger for the pink
colour to disappear. As it
starts taking longer for
the pink colour to disappear
, the sodium
hydroxide is added a dro
p at a time.
Acid-Base Titration
1
reached
• The equivalence point of the titration isonium and
when equal numbers of moles of hydr
hydroxide ions have been reacted.
of the
• When this happens in this titration, the pH
the
and
7.0
is
flask
the
in
tion
solu
phenolphthalein indicator is colourless.
possible.
• Step 4: Add as little excess NaOH as to the
NaOH
of
drop
le
sing
a
We want to add
the
colourless solution in the flask and have
while
solution in the flask turn pink and stay pink
led.
swir
are
flask
the
of
ents
cont
the
ator is
• This permanent colour change in the indic
the
known as the endpoint of the titration and
.
over
titration is
To Solve the problem
write the equation for the reaction:
HCI(aq) + NaOH(aq) —* NaCI(aq) + H20(l)
• lt
•
d
2
n
solve for the amount of moles of the
N8OH mcI
titrant used,
NaOH
0150 molIL x 0.02567 Lo 3.85x 10-3 mol
/
3
using stoichiometry, solve for the
•
mole
concentration of HCI knowing it is a 1:1
Foi.md wi titration experiment
Tftraon
100oL
I
B
OOI
LMC
C
1
2
0
0.06
0.1
015
6.2
Moles NGH u1dod
SAMPLE PROBLEM 2
• In an acid-base titration, 17.45 mL of
HNO were
,
0.180 M nitric acid, 3
completely neutralized by 14.76 mL of
Al(OH) Calculate
.
aluminium hydroxide, 3
aluminium
the
of
the concentration
e.
hydroxid
,
ratio
3.85 x 10-3 mol= 0.192 M
002000 L
SAMPLE ANSWER 2
:
The balanced equation for the reaction
)
3HN03(aq) + Al(OH)3(aq) — A1(N03)3(aq
+
3)120(i)
is:
The number of moles of nitric acid used
x 10-3 mcI HNO3
y mci = 0.180 mci/I x 0.017451 = 3.14
number of moles of
From the stoichiometry of the reaction, the
ah,imwuum hydroxide reacted is:
1.05x 10-3mol
)3=
3,14x 10-3molHNO3x 1 mc AliOH
3 mcI 8N03
tum hydroxide is:
Therefore, the concentration of the akimin
1.05 x 10-3 mcI AtOH)3 0.0711 M
0.01476 1
2