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Precision, Accuracy, and Precipitation Reactions
29 March 2024
CHEM 1211K Laboratory
Data and Results
Unknown letter:
A
Unknown + sodium sulfate
Unknown + lead(II) nitrate
Clear after unknown added and no precipitate formed
Clear after unknown added
and no precipitate formed
Unknown + sodium hydroxide
Unknown + sodium
chloride
Turned light blue in color when unknown added and precipitate
formed
Clear after unknown added
and no precipitate formed
Unknown + barium nitrate
Clear after unknown added and no precipitate formed
Identity of unknown solute:
Copper(II) Nitrate
Table 1.
Determination of the identity of an unknown solution of an ionic salt via precipitation reactions.
Masses of water delivered (g)
Instrument
Trial 1
Trial 2
Trial 3
Mean
St. dev.
Serological pipet
10.144
9.76
10.118
10.007
0.215
Volumetric pipet
10.147
10.185
10.275
10.202
0.066
Graduated cylinder
9.566
10.118
9.83
9.838
0.276
Most accurate:
Serological Pipet
Most precise:
Volumetric Pipet
Table 2.
Determination of precision and accuracy of laboratory glassware designed to deliver 10 mL of liquid.
0
100
200
300
400
500
600
20
30
40
50
60
70
80
90
100
f(x) = 0.11 x + 33.18
R² = 0.95
Time (s)
Temperature (°C)
Figure 1.
Graph depicting the heating of 50mL of water over 600 seconds.
Discussion
Experiment A involved determining the chemical formula of an unknown salt solution, my group chose Unknown A, by creating precipitates with other know solutions. Through the mixing of Unknown A with various solutions and the formation of a precipitate, I have found Unknown A to be the compound copper(II) nitrate. It was observed that sodium hydroxide formed a precipitate with Unknown A and was the only known solution to do so.
Cu
¿
When given the possible cations of Cu
2+
, Ba
2+
, Ag
+
, or Na
+
along with the possible anions of Cl
–
, NO
3
–
, or SO
4
2–
; a total of 12 different salt solutions can be made. Using solubility rules however helps narrow down the selection to just a single salt solution. Any salt solutions containing sodium were eliminated because salts of group 1 cations are soluble and would therefore not create a precipitate with sodium hydroxide. Salt solutions containing silver could be ruled out since a sulfate becomes insoluble when combined with silver, thus a precipitate would have been observed when mixing Unknown A with sodium sulfate which was not the case. The same explanation can be made with the use of barium as well, leaving only the copper salt solutions. Copper(II) chloride would have formed a precipitate with lead(II) nitrate and copper(II) sulfate would have formed a precipitate with barium nitrate, both of which were not observed in actual testing. Simple process of elimination leads to the only possible salt solution of copper(II) nitrate to be the Unknown A solution in Table 1.
Table 2 gives data on experiment 2 where we were tasked to measure out 10 mL
of water with three different pieces of glassware. The glassware used was a graduated cylinder, a volumetric pipet, and a serological pipet in order to test their accuracy and precision. After collecting our data we observed that the accuracy of the glassware from
most accurate being the serological pipet, then the graduated cylinder, and the least accurate being the volumetric pipet. The order of accuracy was determined by the percent error, which shows how close each tool actually got to the 10 mL of water. Percent error can be calculated using the formula:
PE
=
x
meas
−
x
true
x
true
×
100
Calculations provide the serological pipet had a 1.18 percent error in trial 3, the graduated cylinder had a 1.18 percent error in trial 2, and the volumetric pipet had a 1.47 percent error in trial 1. Although the serological pipet and the graduated cylinder have equal percent errors, the next closest in accuracy was the serological pipet with a percent error of 1.47 in trial 1, while the graduated cylinder only had a percent error of 1.70 in trial 3 making the serological pipet more accurate. However when it comes to precision, the volumetric pipet was the most precise, then the serological pipet, and the least precise was the graduated cylinder. Precision shows how consistent the glassware
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Related Questions
Does the number of ions in solution increase, decrease, or remain constant?
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Table 1: Qualitative Analysis of Group II Cations
Group II
Cations
Confirmatory
Reagent
Result/
Observation
Substance
Formed
Inference
Write the chemical equations involve in the analysis of group II cations.
Guide Question:
What are differences of the precipitates of each cation in group II?
Reference video : https://www.youtube.com/watch?v=BZGZZx1lHKA
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The Solubility Product Constant for zinc phosphate is 9.1 x 10-33
Solid sodium phosphate is slowly added to 125 mL of a 0.0500 M zinc acetate solution. The concentration of phosphate ion required
to just initiate precipitation is
M.
Submit Answer
Retry Entire Group 7 more group attempts remaining
Submit Answer
Use the References to access important values Here
The Solubility Product Constant for copper(11) hydroxide is 1.6 x 10-19,
Solid sodium hydroxide is slowly added to 50.0 mL of a 0.0340 M copper(II) nitrate solution. The concentration of hydroxide ion
M.
required to just initiate precipitation is
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Use the References to access important values if needed for this question.
The Solubility Product Constant for calcium fluoride is 3.9 × 10-11.
The maximum amount of calcium fluoride that will dissolve in a 0.102 M potassium fluoride solution is
Submit Answer
Retry Entire Group 8 more group attempts remaining
Fil
When 25.0 mL of a
1.78 x 10-4 M barium chloride solution is combined with 15.0 mL of a
3.67 x 10-4 M potassium carbonate solution does a precipitate form? (
Kap (BaCO3) = 8.1 × 10-⁹)
OYes, the precipitate forms.
ONo, the precipitate doesn't form.
For these conditions the Reaction Quotient, Q, is equal to
When 25.0 mL of a
9.96 x 10-4 M sodium bromide solution is combined with 15.0 mL of a
1.82 x 10-4 M lead nitrate solution does a precipitate form? (
Ksp (PbBr₂) = 6.3 × 10-8)
M.
OYes, the precipitate forms.
ONO, the precipitate doesn't form.
For these conditions the Reaction Quotient, Q, is equal to
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Procedure 2:
• You add cold HCI and notice a white precipitate.
• You set the test tube in boiling water and you notice you still have precipitate. You
decant the liquid into a different test tube and keep both
Procedure 3:
• You split the decanted solution in half, into two different test tubes
• In the first half, you add potassium chromate into the solution. You notice a yellow
precipitate
• In the second half, you add sulfuric acid. You notice a white precipitate
Procedure 4:
• You add ammonia into the test tube. The test tube off-gasses
• You then centrifuge the tube. You notice there is no precipitate
• To the solution, you add nitric acid until the solution is acidic (which you test with
blue litmus paper)
• You notice a white precipitate
Cations Present, Undetermined, or Absent:
Pb: ?
Hg: ?
Ag: ?
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only SELECT correct respomses
1) Titrations of different forms are often employed to quantitate acids, bases, metal ions, and proteins.
2) To perform a titration, the experimenter must be familiar with the stoichiometry of the reaction taking place between the reagent and analyte.
3) Often in titrations, a visible color change will occur in the solution (due to an added indicator solution).
4) Gravimetric titrations directly measure the volume of reagent solution needed to react completely with an analyte.
5) For a monoprotic weak acid titration, the weak acid solution being titrated is the "sample" that is being quantitated.
6) For a monoprotic weak acid titration, the equivalence point is reached when the moles of added base equals the moles of weak acid originally present in the reaction vessel.
7) In a monoprotic weak acid titration, the weak acid is the analyte in solution, and a standardized base (for example, a solution of known concentration of NaOH) is the titrant.
8) A…
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Please help...
Please provide only typed answer solution no handwritten solution needed allowed
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*WRITE RESULTS AND DISCUSSIONS OF TITRATION- CONCENTRATION OF VINEGAR
*WRITE CONCLUSION
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Solid ammonium phosphate is added slowly to this mixture.
What is the concentration of chromium(III) ion when calcium ion begins to precipitate?
[Cr³+]
Submit Answer
M
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A solution contains 1.02x10-2 M zinc nitrate and 1.21x10-2 M silver acetate.
Solid potassium cyanide is added slowly to this mixture.
What is the concentration of silver ion when zinc ion begins to precipitate?
[Ag+] =
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please answer in text form and in proper format answer with must explanation , calculation for each part and steps clearly
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Quantity
Your Data
1.
Grams of vinegar sample used for your titration
25.000 g
2.
Initial Buret Reading of Sodium Hydroxide solution
10.00 mL
3.
Final Buret Reading of Sodium Hydroxide solution
28.00 mL
4.
Amount of Sodium Hydroxide Solution used to neutralize the vinegar sample
5.
Concentration of NaOH in the NaOH solution
0.050 g/mL
6.
Grams of NaOH used to neutralize the vinegar
7.
Grams of acetic neutralized by the amount of NaOH
8.
Percent acetic acid in the vinegar
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For #6, Once you calculate the amount of sodium hydroxide used, multiple that value by the concentration of NaOH in the NaOH solution.
For #7, Multiply the value obtained in number 6 by the number 1.5. Remember, we learned that
every 1 gram of NaOH neutralizes 1.5 grams of acetic acid.
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ollowing combinations would NOT form a precipitate in aqueous
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-) Sr(NO:)2 and NaOH
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Zn(C:H3O2)2 and Na:S
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[hydroxide] =
M
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CHE 1560- General Chemistry II Laboratory
Experiment 8: Solubility Product Make Up Assignment
BEAKER 2
BEAKER 3
M of NH3 solution
5 M
Initial reading of buret
0 mL
10.4 mL
(NH3) (mL)
Final reading of buret (NH3) 10.4 mL
(mL)
19.8 mL
Total volume of solution
30.4 mL
29.4 mL
after titration
Total moles of C2042-
Molarity of C2042-
Total moles of Cd2+
Moles of [Cd(NH3)42+]
Molarity of [Cd(NH3)42+]
Moles of NH3 added by
titration
Moles of NH3 that did not
react with Cd2+
Molarity of NH3 that did not
react with Cd2+
Kr for [Cd(NH3)42+]
Average K for [Cd(NH3)42+]
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Review Homework REQUIRED (NOT Pie Progress)
Question 16 of 46 (2 points) | Question Attempt: 1 of Unlimited
✓ 14
15
16
17
18
19
20
21
✓ 22
✓ 23
Complete the table below by deciding whether a precipitate forms when aqueous solutions A and B are mixed. If a precipitate will form, enter its empirical
formula in the last column.
solution A
solution B
Does a
precipitate form
when A and B
are mixed?
empirical
formula of
precipitate
Do
ammonium sulfide
copper(II) nitrate
Oyes no
☐
silver nitrate
sodium carbonate
O yes no
potassium chloride
barium nitrate
O yes no
☐
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References
Use the References to access important values if needed for this question.
A solution contains 7.78x10-3 M sodium hydroxide and 1.44x10-² M sodium
cyanide.
Solid zinc acetate is added slowly to this mixture.
What is the concentration of hydroxide ion when cyanide ion begins to precipitate?
[hydroxide] =
M
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3
Solid potassium fluoride is slowly added to 75.0 mL of a 0.233 M calcium bromide solution until the concentration of
%.
fluoride ion is 0.0632 M. The percent of calcium ion remaining in solution is
Submit Answer
20
E
$
4
R
F
Use the References to access important values if needed for this question.
Retry Entire Group
%
5
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2.15.0 mL HNO3 was titrated to equivalence with 60.0 mL of 0.020
NaOH. What is the concentration of the acid?
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A solution contains 7.39x10-³ M sodium carbonate and 6.78x10-3 M sodium phosphate.
Solid calcium nitrate is added slowly to this mixture.
What is the concentration of phosphate ion when carbonate ion begins to precipitate?
[phosphate] =
M
Submit Answer
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HC2H3O2 (aq) + NaOH (aq) ----> NaC2H3O2 (aq) + H2O(l)
Trial
Vi (mL)
Vf (mL)
1
2.25
15.76
2
3
For their first trial, a student pipettes 5.00 mL of a vinegar solution into an Erlenmeyer flask. The student titrates the vinegar with standardized NaOH with a molarity of 0.317 M. The volume of titrant used was recorded in the table above. Calculate the molarity of the vinegar solution for this trial. Enter a number without units.
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Please answer all parts
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ted
Solid manganese (II) sulfate is slowly added to 75.0 mL of a potassium carbonate solution until the concentration
of manganese(II) ion is 0.0181 M. The maximum amount of carbonate remaining in solution is
M.
80
Submit Answer
$
4
R
V
%
5
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Retry Entire Group 5 more group attempts remaining
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