So I know that some may come across this and think to yourselves that I am just

Question

So I know that some may come across this and think to yourselves that I am just a lazy student looking for answers. I am truly not looking for the answers... If you could provide them then that would be great but I am just looking for the formula of how to do them myself, so I can learn. I have a tough time with this subject and it is my first time taking it since highschool which was 10 years ago. So if you could help; I am grateful

Molarity of standard NaOH solution: x/1 H 2/3 Unknown + 2/3 NaOH -> 2/3 H2O +Na 2/3 Acetate

Unknown Acid code and number of reactive hydrogens: H+3

1. Find the volumes of acid solution (vinegar) and base solution used in each of your vinegar titrations.

2. Find the moles of base (NaOH) used in each of your vinegar titrations.

3. Using stoichiometry and the balanced equation, find the moles of acetic acid used in each of your vinegar titrations.

4. Find the molarity of the vinegar for each of your three vinegar titrations, then find the average molarity of the acid.

5. Find the percent deviations for your three vinegar molarities for the vinegar titrations.

6. From the moles of acetic acid for each vinegar titration, find the mass of acetic acid for that titration

7. Assuming that the density of vinegar is the same as the density of water, find the mass of vinegar for each titration, and calculate the mass percent of acetic acid forthat titration. (You looked up the density of water back in the measurement lab.)

8. Average your three mass percents of acetic acid to get your final answer for the

mass percent of acetic acid.

9. Find the volumes of base solution for each of your unknown acid titrations.

10. Find the moles of base (NaOH) for each of your unknown acid titrations.

11. Find the balanced equation using the number of reactive hydrogens (call an acid with one reactive hydrogen HA, one with two reactive hydrogens H2A, and so on).

12. Find the moles of unknown acid for each of your three unknown acid solutions.

13. Use the mass of unknown acid and the moles of unknown acid to find the formula mass of your unknown acid for each of your three unknown acid titrations.

14. Average your three formula masses to get your final answer for the formula mass of

your unknown acid.

15. Calculate the percent deviation for each of your three formula masses.

16. AgBr reacts with Na2S2O3 according to the following equation:

            AgBr (s) + 2 Na2S2O3 (aq) ==> Na3Ag(S2O3)2 (aq) + NaBr (aq)

        How many milliliters of 0.0138 M Na2S2O3 are needed to dissolve 0.250 g of AgBr?

17. Write the balanced double displacement reaction for Pb(NO3)2 reacting with NaCl.

How many milliliters of 0.750 M Pb(NO3)2 are required to react completely with 525 mL of 2.25 M NaCl?

Vinegar: Titration 1 Titration 2 Tirtration 3 Initial V of Acid .01 mL 5.56 mL 11.32 mL Final V of Acid 5.56 mL 10.71 mL 16.02 mL Inital V of Base .01 mL 27.12 mL 3.01 mL Final V of Base 22.23 mL 44.71 mL 24.90 mL

Explanation / Answer

Using the following formulas you can do the calculation yourself.

1. Volume of acid solution or vinegar used = Final volume - Initial volume

For example, for titration 1: Volume of acid = 5.56 - 0.01 = 5.55 mL

Volume of base used = Final volume - Initial volume

For example, for titration 1: Volume of base = 22.23 - 0.01 = 22.22 mL

2. I think you are supposed to determine the concentration of acetic acid by standardising it with

a known concentration of NaOH solution. You know the strenght of NaOH solution. Let it be S1 (M).

So 1000 mL of NaOH solution contains S1 mol of NaOH.

Then, 22.22 mL NaOH solution contains (S1 * 22.22 /1000) mol of NaOH.

So, mol of base used for 1st titration = (S1 * 22.22 /1000) mol

3. CH3COOH + NaOH = H2O + CH3COONa

So, 1 mol of base required to neutralise 1 mol of acid.

So no of moles of base = no of moles of acid

4. You can calculate molarity of vinegar(S2) according to the equation, V1 S1 = V2 S2

V1= volume of base used

S1= Strength of NaOH solution (which is known)

V2= Volume of acid used

S2= Unknown molarity or strength of Vinegar

5. Let the literature value of molarity of Acetic acid in vinegar is X.

Then % deviation = (X - S2 / X) * 100

6. You know the mol of acid used (From 3).Then mass of acid

= mol of Acid * Molar mass of acetic acid = (mol of acetic acid * 60) g

7. Mass of vinegar = Density of Vinegar * Volume of vinegar ( From 1)

mass percent of acetic acid = (mass of acid/ mass of vinegar) * 100

8. Average the three values (From 7) for 3 titrations.

9. volumes of base solution = Final volume - Initial volume

10. Same as question 2.

11. H3A + 3NaOH = Na3A + 3 H2O

12. 3 mols of base is required to neutralise 1 mol of acid. You know the mol of base (From 10).

So, mol of acid = mol of base / 3

13. Formula mass of acid = Mass of acid required to prepare 1 molar acid solution.

You know the mass of acid (a) and mol of acid(M) used.

So, Formula mass(f) = (a/M) gram

14. Average the above three values.

15. Let the literature value of formula mass of unknown acid is Y.

Then % deviation = (Y-f/Y) * 100

16. AgBr (s) + 2 Na2S2O3 (aq) ==> Na3Ag(S2O3)2 (aq) + NaBr (aq)

0.250 g of AgBr = 0.250/187.77 mol= 0.001 mol

2 mol of thiosulfate is required to dissolve 1 mol of AgBr.

So, 0.002 mol of thiosulfate is required to dissolve 0.001 mol of AgBr.

0.002 mol of thiosulfate = 0.002* 158.11 = 0.316 g of thiosulfate

Now, 1000 mL of 0.0138 M Na2S2O3 = 0.0138 * 158.11 = 2.18 g Na2S2O3

So, Milliliters of 0.0138 M Na2S2O3 are needed to dissolve 0.250 g of AgBr

= 1000 * 0.316 / 2.18

= 145 mL

17. Pb(NO3)2 + 2 NaCl = PbCl2 + 2NaNO3

525 mL of 2.25 M NaCl

= 2.25 * 58.44 * 525/1000

= 69 g NaCl

116.88 g NaCl reacts with 331.21 lead nitrate. So, 69 g will react with = 331.21*69/116.88= 195.53 g Lead nitrate.

1000 mL of 0.750 M Pb(NO3)2 = 331.21*0.750 = 248.4 g

Milliliters of 0.750 M Pb(NO3)2 are required = 1000*195.53/248.4 =787 mL

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