Titration Experiment

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Titration Expt. Overview
• Day 1: You will create a ~0.1 M NaOH solution
• Day 2: You will standardize your solution of
NaOH by titrating it with a solid acid, KHP
• Purpose: To verify your [NaOH] up to 3 SF
• It should be close to 0.100 M NaOH
• Day 3: You will titrate your now known
concentration of NaOH with an unknown
concentration of HCl
• Purpose: To determine the [HCl]
Day 1: Making a NaOH solution
• You and your partner will make 250 mL of a ~0.1
M NaOH solution
• How will you do this? What do you need to
know? What equipment do you need?
• Remember, M = mol/L
• Measure NaOH mass using a balance and dissolve
in 250.0 mL distilled water together in a
volumetric flask
• Calculate and record exact [NaOH] in M (mol/L)
• 3 SF
• Should be close to 0.100 M, but won’t be exact
Day 2: Standardization of NaOH(aq)
• Purpose: To verify your [NaOH] by reacting
(titrating) it with an known acid, KHP
(PHTH)
Buret w/ base
~0.1 M NaOH
Flask w/ acid
and PHTH ind.
~0.3 g KHP in
~75 mL H2O
Faint pink
tinge…
Use white paper
underneath to
see better
Day 2: Standardization of NaOH(aq)
• Purpose: To verify your [NaOH] by reacting
(titrating) it with an known acid, KHP
To neutralize… mol H+ = mol OH• Acid = ~0.3 g solid KHP (potassium hydrogen phthlate)
• mass g KHC8H4O4  mol KHP = mol H+
• Recall M = mol/L, so mol OH- = M OH· L OH
therefore… mol H+ = M OH · L OH
• Do one dry run  ball park amount NaOH
• Do at least two good trials, if not three
• Average your [NaOH] results
Day 3: Titration of HCl(aq) with NaOH(aq)
• Purpose: To determine unknown [HCl] by
reacting (titrating) it with known base, NaOH
Buret
w/ base
~0.1 M
NaOH
Flask w/
acid and
ind.
Yellow
~ 15 mL of
? M HCl
Green
Endpoint
Use white paper
underneath to
see better
Blue
Overshoot
(too basic)
Day 2: Standardization of NaOH(aq)
• Purpose: To verify your [NaOH] by reacting
(titrating) it with an known acid, KHP
To neutralize… mol H+ = mol OH• Acid = ~0.3 g solid KHP (potassium hydrogen phthlate)
• mass g KHC8H4O4  mol KHP = mol H+
• Recall M = mol/L, so mol OH- = M OH· L OH
therefore… mol H+ = M OH · L OH
• Do one dry run  ball park amount NaOH
• Do at least two good trials, if not three
• Average your [NaOH] results
Lab Report
• Typed (calculations may be hand-written)
• Sections
Written portions should be:
• Overview (3-5 sentences)
• Procedure
• 3rd person
• Passive voice
• Past tense
• Day 1: Making NaOH solution
• Day 2: Titration (Standardization) with KHP
• Day 3: Titration with HCl
• Data Tables
• Show organized data for all three days of expt. (include dry
run data and label as such)
• Include related data table with each day’s procedure
• Calculations
• Show all calculations (no need to use dry run data)
• Include related calculations with each day’s procedure
strong acid
HCl (aq)
+
[H+] = 0.50 M
~ 15 -20 mL
+ strong base  salt + HOH
product pH = 7… use bromothymol blue
yellow (acid), green (neutral), blue (base)
NaOH (aq)
[OH-] = ? M
VB
 NaCl (aq) +
HOH (l)
Measure both V in expt. (to the nearest 0.00 mL)
Trials
1. (Dry Run)… calculations unimportant (ball park numbers)
2.
3.
4.
Calculate [OH-] for trials 2-4 and then average
Strong Acid/Strong Base Titration
HCl + NaOH  H2O + NaCl
Titrate with Bromthymol Blue
pH
At this point,
everything is water,
Na+ and Cl-, so the
pH = 7
Equivalence Pt
mol H+ = mol OH-
Overshoot
Endpoint
mL of base
Weak Acid/Strong Base Titration
+ strong base  weak base + HOH
product pH > 7… use phenolphthalein (PHTH)
clear  acid
pink  base
HCH3COO (aq) + NaOH (aq)  NaCH3COO (aq) + HOH (l)
(acetic acid,
0.24 M
(Na+ (aq) +
i.e. vinegar)
VB
CH3COO-(aq))
[H+] = ?
~ 3-5 mL
Weak acid
Trials
1. (Dry Run)… calculations unimportant (ball park numbers)
2.
3.
4.
Weak Acid/Strong Base Titration
HC2H3O2 + NaOH  H2O + NaC2H3O2
Titrate with Phenolphthalein
At this point, the
solution contains water,
Na+, and C2H3O2 - , so
the pH > 7.
pH
BUFFERING REGION
HC2H3O2
C2H3O2 - + H+
mL of base
Equivalence Pt
mol H+ = mol OH-
Endpoint
Overshoot
Weak Base/Strong Acid Titration
HCl + NH4OH  H2O + NH4Cl
Titrate with Methyl Orange
NH3 + H+
NH4+
BUFFERING REGION
pH
At this point, the
solution contains
water, Cl-, and NH4+,
so the pH < 7.
Equivalence Pt
mol H+ = mol OH-
Endpoint
Overshoot
mL of acid
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