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18 Dec 2019

A saturated aqueous solution of a slightly soluble salt, bariumiodate [Ba(IO3)2], was prepared and titrated with sodium thiosulfate. Based on the data tabulated below, calculate the indicated quantities in the questions.

Volume of saturated barium iodate soution: Initial Before titration, volume of thiosulfate in the burette: Final (after titration) volume of thiosulfate in the burett: Molarity of Thiosulfate sln

1.0 ML 0.0 ML 6.5 ML 0.01 mol/L

2.0 ML 6.8 ML 20.1 ML 0.01 mol/L

3.0 ML 20.3 ML 46.8 ML 0.01 mol/L

Organize your calculations using the steps below, but first determine and write the proper chemical equation

A. Calculate the number of moles of thiosulfate used in each titration: n IO3 - =

B. Calculate the umber of moles of iodate tirated: n IO3 - =

C. Calcuate the molar concentration of iodate ions: [IO3 -]

D. Calculate the average molar concentration of iodate ions: Avg[IO3-] =

E. Calculate Ksp for the dissolution of Ba(IO3)2: Ksp =

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Nestor RutherfordLv2

31 Dec 2019

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Please check through my work, not 100% sure I took the rightsteps to doing this or not.

A saturated aqueous solution of slightly soluble salt, bariumiodate [ Ba(IO3)2] was prepared and titrated with sodiumthiosulfate. Based on the data tabulated below, calculate theindicated quantities in questions 1-5.

(Column A)Volume of saturated Barium Iodate solution (filteredsolution).
(Column B) Initial (before titration ) volume of the thiosulfate inthe burette.
(Column C) Final (after titration) volume of the thiosulfate in theburette.
(Column D) Molarity of titrant (thiosulfate solution). 1,2,3 aretests.
------A------B--------C---------D
---[ mL ]--[mL]--- [mL]---[mol/L]
1) [ 1.0 ]--[0.0]---[6.5]---[0.01]
2) [ 2.0 ]--[6.8]---[20.1]--[0.01 ]
3) [ 3.0 ]--[20.3]- [46.8]--[0.01]

Organize your calculations using the steps below, but firstdetermine and write the proper chemical equation. Ba(IO3)2 +Na2S2O3---> ??

1mol IO3^- reacts with 6mol S2O3^2-

1. Calculate the number of moles of thiosulfate used in eachtitration: n S2O3^2- =
1. 0.01x0.00 = 0
2. 0.01x0.0068 = 0.000068 = 6.8e^-5
3. 0.01x0.0203 = 0.000203 = 2.03e^-4
2. Calculate the number of moles of Iodate titrated: n IO3^-=
1. 0/6 = 0
2. 6.8e^5/6 = 0.0000113 = 1.13e^-5
3. 2.03e^-4/6 = 0.0000338 = 3.38e^-5

3. Calculate the molar concentartion of Iodate ions: [IO3^2-]
1. 0/0.001 = 0
2. 1.13e^-5/0.002 = 0.00565 = 5.65e^-3
3. 3.38e^-5/0.003 = 0.01126 = 1.13e^-2

4. Calculate the average molar concentration of Iodate ions: Avg[IO3^2-] =
0 + 0.00565 + 0.01226 = 0.01695/3 0.00565 = 5.65e^-3

5. Calculate Ksp for the dissolution of Ba (IO3)2 : Ksp =

Ksp = [Ba^2+][IO3^-]^2

blackjellyfish847

1.) average number of moles of NaOH dispensed

2.)average concentration of the acid (mol/L)

3,) Based on your coarse titration volume, do you expect the acetic acid solution to have a higher or lower concentration than the NaOH solution?

higher could you please tell me if number 3 is right

please help with these 2 questions

Part 1: Prepare the NaOH Solution

Lab Results

How many mL of water were used to prepare the NaOH solution?

200mL

Data Analysis

Calculate the molarity of the NaOH solution. The molar mass of NaOH is 39.997 g/mol.

0.1M

Calculate the amount of benzoic acid to be neutralized by 20.00 mL NaOH solution, in both moles and grams. The molar mass of benzoic acid is 122.12 g/mol.

1.207 g

Part 2: Perform a Coarse Titration

Lab Results

Record the following data from your course titration in the table below.

mass of benzoic acid used (g)	0.26614g
pH of benzoic acid solution before titration	2.96
volume of NaOH in the burette before titration (mL)	50.0mL
volume of NaOH in the burette after titration (mL)	28.89mL
volume of NaOH dispensed in the titration (mL)	21.11mL

Data Analysis

How do you expect your coarse titration volume to compare to your fine titration volumes?

I expect the values to be a little off from each other since in the first titration we let out a lot at a time and didnt slowly do the titration. should only be off by a few mL.

Part 3: Perform Fine Titrations

Lab Results

Record the volume of NaOH solution dispensed in the 3 fine titrations.

20 mL dispenced ,20.34mL, 20.30mL

Data Analysis

Calculate the average concentration of the NaOH solution, using the average volume of NaOH solution dispensed in the 3 fine titrations. Report your answer using enough significant figures to distinguish it from the expected concentration of 0.100 M.

V = (V₁ + V₂ + V₃) / 3

V = (20.03 mL + 20.06 mL + 20.02 mL) / 3 = 20.04 mL

M = n / V = 0.00200 moles / 0.02004 L = 0.0998 M

Learning objective: Determine the concentration of a solution.

20.21= ----> 0.0999M

Experiment 2: Use the Standardized NaOH Solution to Determine the Concentration of an Acid

Part 1: Perform a Coarse Titration

Lab Results

What was the pH at the end point of the coarse titration?

3.87ph

Data Analysis

Based on your coarse titration volume, do you expect the acetic acid solution to have a higher or lower concentration than the NaOH solution?

higher

Part 2: Perform Fine Titrations

Lab Results

For the 3 fine titrations of the acid of unknown concentration, fill in the following data.

	Titration #1	Titration #2	Titration #3
volume of acid (mL)	59.11	40.97	40.98
volume of NaOH dispensed (mL)	9.05	9.03	9.02

Data Analysis

Calculate the following quantities and record the data in the table below.

average volume of NaOH solution dispensed (mL)	9.03
average number of moles of NaOH dispensed
average concentration of the acid (mol/L)

Conclusions

Phenolphthalein is pink over the range of pH 8 â 12. Why was it a useful indicator of when the equivalence point was reached?

This was helpful because once I got to the ph of 7 I knew that I would be getting closer to the end point which would change the color to pink.

Suppose a student titrated a sample of monoprotic acid of unknown concentration using a previously standardized solution of NaOH.

volume of 0.125 M NaOH dispensed	24.68 mL
volume of acid solution	50.00 mL

Given the data in the figure above, what is the concentration of the unknown acid?

0.0617M

Lab NOTES:

ex1

Take sodium hydroxide from the Materials shelf and add 0.800 g to the beaker.

Take water from the Materials shelf and add 200 mL to the beaker.

Ph 2.96

inital volume in burett

inital volume 50ml

1.) volume43.78mL

volumes dispenced =6.22ml

2. volume =38.05

volume dispenced =11.95ml

3.)volume = 32.06

volume dispenced=17.94

4.28.89ml

volume dispenced=21.11ml

ph=10.95

--------------------------------------

Ex 2

inital ph= 2.96

volume30.33

dispence 20.05

1. volume 39.92ml

volume dispenced=10.08

2.volume 29.66

voiume dispence 20.34

final ph10.43

initial valume in burett 50ml

--------------------

ex3

final ph9.43

volume 30.28

volume dipenced=20.00

----------------------------------------------------------------------

2nd Ex

initial ph 3.11

volume 50

1. volume 48.95mL

dispenced 1.05mL

ph 3.87

2. volume 45.88

dispenced 4.12

ph 4.67

3 volume 42.99

dispenced 7.01

ph 5.29

4 volume 39.93

dispence 10.07

ph 11.45

--------------------

final ph 9.99

volume 59.11

dispenced 9.05

ph 9.59

volume 40.97

dispenced 9.03

ph 9.50

volume 40.98

9.02

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