Mastering Acid-Base Titrations: Equivalence Points & Curves
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E
6-14: Study of Acid-Base Titrations – Polyprotic Acids
In this video we'll show you how to operate the buret:
Titrations provide a method of quantitatively measuring the concentration of an unknown solution. In an acid-base titration, this is done by delivering a titrant of known
concentration into an analyte of known volume. (The concentration of an unknown titrant can also be determined by titration with an analyte of known concentration and
volume.) Titration curves (graphs of volume vs. pH) have characteristic shapes. The graph can be used to determine the strength or weakness of an acid or base. The
equivalence point of the titration, or the point where the analyte has been completely consumed by the titrant, is identified by the point where the pH changes rapidly over a
small volume of titrant delivered. For polyprotic acids, there will be multiple equivalence points. In this assignment, you will observe this titration curve by titrating the weak
acid H
2
SO
3
with the strong base NaOH.
1. To start this activity, click this link for Study of Acid-Base Titrations – Polyprotic Acids
. The lab will load in a new tab. Click back to this tab to read further instructions
and complete the questions below.
2. The buret will be filled with 0.3106 M NaOH and 20.00 mL of 0.2556 M H
2
SO
3
will be in the beaker with thymol blue as an indicator. In the Buret section of Live Data
click the button that says Save Data
. This ensures that all of your titration data is saved to the Lab Book. The horizontal position of the orange handle is off for the
stopcock. Open the stopcock by pulling down on the orange handle. The vertical position delivers solution the fastest with three intermediate rates in between. Turn the
stopcock to one of the fastest positions. Observe the titration curve. When the volume reaches 40 mL, double-click the stopcock to stop the titration. Click Stop
Saving
and the data will appear in a table in the Lab Book.
Write complete balanced equations for the two-step neutralization reaction between H
2
SO
3
and NaOH.
00:53
Here are the balanced equations for the neutralization reaction between sulfuric acid (H2SO3) and sodium hydroxide (NaOH) in two steps:
Step 1: H2SO3 + NaOH →
NaHSO3 + H2O
This step involves the neutralization of one hydrogen ion from sulfuric acid with one hydroxide ion from sodium hydroxide, forming sodium bisulfite and water.
Step 2:
NaHSO3 + NaOH →
Na2SO3 + H2O
In this step, the remaining hydrogen ion from sodium bisulfite reacts with another hydroxide ion from sodium hydroxide, forming sodium sulfite and water.
Edit Insert Formats
P
3.
What was the pH and color of the solution at the beginning of the titration?
pH = 1.30
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Beyond Labz 2.0 Assessment
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Color =
Red
Clear
Orange
Yellow
Blue
Green
4.
What was the pH and color of the solution at the end of the titration?
pH = 12.40
Color =
Clear
Blue
Green
Red
Yellow
5. Did any additional color changes occur during the titration?
During the titration, two other color changes were observed which were yellow and green.
6. Examine the graph of pH vs. volume (blue line) and attach a sketch or screenshot of the titration curve below.
Screenshot 2… 3.12.10PM
Choose File
7. What happens to the pH around 16 mL and 32 mL? What causes each to occur?
In a titration, pH rapidly changes around 16 mL and 32 mL, indicated at two points, equivalence points. At 16 mL, the first hydrogen ion of sulfuric acid is neutralized, and at 32 mL, the second ion. The rapid pH changes
because the titrant and analyte react completely in stoichiometric proportions.
8.
Examine the graph of conductivity vs. volume (red line).
What happens to the conductivity during the titration?
As the titration begins, the conductivity initially remains stable at around 23.2 µ
S/cm. However, it gradually decreases until approximately 16 mL, when it increases rapidly. It reaches a maximum of around 32 mL, then st
decrease again. This behavior can be explained as follows:
At the beginning of the titration, the acid (H2SO3) and the base (NaOH) are in low concentrations. Therefore, the conductivity remains stable.
As the strong base (NaOH) is added to the weak acid (H2SO3), neutralization reactions lead to the formation of water and salts. These salts initially decrease conductivity.
At around 16 mL, the pH increases rapidly, indicating the nearing equivalence point for the titration of the first acidic hydrogen ion of sulfuric acid. At this point, excess hydroxide ions from NaOH contribute to a significa
increase in ionic concentration.
The conductivity reaches its maximum around 32 mL, corresponding to the equivalence point for titration of sulfuric acid's second acidic hydrogen ion. At this point, all the sulfuric acid is neutralized, resulting in the ma
concentration of ions in solution. Beyond the equivalence point, further addition of the base leads to excess hydroxide ions, which can slightly decrease conductivity.
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Concentration sodium thiosulfate solution used:
0.056 mol L-1
ACCURATE TITRATION VOLUME OF SODIUM THIOSULFATE SOLUTION USED
Volume titration 1, (ml)
Volume titration 2, (ml)
Your Group
15.80
15.75
Group 1
16.05
15.70
Group 2
16.30
16.25
ANALYSIS
Average volume used, (mL)
Standard deviation
Number of moles sodium thiosulfate used, (mol)
Number of moles calcium iodate in solution, (mol)
Volume of calcium iodate used, (mL)
Concentration of saturated calcium iodate solution (mol L-¹)
Unrounded value
15.9750
0.26220
Rounded value
15.97
0.262
10.0
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grammarly-G X
*
Content
oard.odu.edu/ultra/courses/_393452_1/cl/outline
F2
Question Completion Status:
X GA student wan X
1
QUESTION 3
A student wants to determine the solubility of water and copper (1) bromide (a green solid) in ethyl acetate (a colorless organic liquid).
She places 1 mL of ethyl acetate into each of two test tubes. To one, she adds 5 drops of water to it. To the other, she adds a small
scoop of copper(1) bromide.
8.0
F3
2
- Which test tube represents the ethyl acetate and
A. Test tube 1
copper(1) bromide if the copper(1) bromide is insoluble in ethyl B. Test tube 2
acetate?
C. Test tube 3
D. Test tube 4
- Which test tube represents the mixture of the ethyl acetate
and water if the water is insoluble in ethyl acetate?
$
Question Com X C Chegg Search x
3
Which test tube represents the mixture of the ethyl acetate
and water if the water is soluble in ethyl acetate?
- Which test tube represents the ethyl acetate and
copper(1) bromide if the copper(1) bromide is soluble…
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Experimental data
Complete the table below.
Trial 1
0.2001 g
Trial 2
0.2065 g
Trial 3
Weight of sodium oxalate
(Na2C2O4, MM= 134 g/mol)
Titration data
0.2050 g
Final reading
Initial reading
29.86 mL
0.00 mL
30.66 mL
30.52 mL
0.00 mL
0.00 mL
Total vol. of KMNO4 used
Computed Molarity of KMNO4
solution
Mean Molarity
Computed Normality of KMNO4
Mean Normality of KMNO4
solution
Reaction Involved:
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Titration Data (Standardization of EDTA)
Example
II
Mass, g Caco,
0.0250
0.0251
0.0246
0.0254
Final reading, mL
25.00
24.91
24.87
25.31
Initial Reading, mL
0.00
0.02
0.02
0.02
Volume, mL EDTA
25.00
Molarity of EDTA, M
0.009992
B. Analysis Data
Example
II
Volume of Sample, mL
25.00
24.90
25.10
25.0
Final reading, mL
27.46
25.08
25.06
24.99
Initial Reading, mL
0.00
0.02
0.02
0.02
Volume of EDTA, mL
27.46
Molarity of EDTA, M
0.009992
Ca" concentration, ppm
440
STATISTICAL EVALUATION OF RESULTS
Standardization of EDTA
Mean, M EDTA
Median, M EDTA
Standard Deviation
Ci for u at 95% Prob. Level
Reported value: M EDTA
Ca" Concentration
Mean, Ca" ppm
Median, Ca" ppm
Standard deviation
Cl for u at 95% Prob. Level
Reported value: Ca" ppm
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The 500.0-mg sample of impure Na2 CO3 (FM: 105.96) required 22.00 mL of the HCI standard solution (the
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ut of
1. Give the balanced chemical equation between the sample, Na2CO3 and the standard, HCI.
2. What is the mass (g) of Na, CO3 in the sample?
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|(mL)
Burette
Burette
Reading
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Trial 1
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??
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HO
HO
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Hard Water Trial 3
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