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# Concentration through Titration

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1. Calculate the concentration of the HCl(aq) solution
( 25mL of .1M unstandardized HCl solution, 2 drops of phenolphthalein indicator used. Titrated with 15.2mL of sodium hydroxide solution)

2. Calculate the concentration of the Na2B4O7 solution two ways
a) From the HCl titration of the Na2B4O7 solution
( 10mL of ~0.07M unstandardized Na2B4O7 solution, 50mL de-ionized water, and 3 drops of methyl red indicator. Titrated with 12.2mL of standardized hydrochloric acid)
b) From the titration with NaOH
 the reaction for the titration of sodium tetraborate with hydrochloric acid solution is: Na2B4O7(aq) + 2HCl(aq) + 5H2O(l)  4H3BO3(aq) + 2NaCl(aq)
(To the solution remaining from (a), added 5 sccops of sugar mannitol, 2 drops of phenolphthalein indicator. Titrated this solution with 15.1mL of standardized NaOH solution.)

3. Determine the theoretical and percent yields and purities for boric acid and sodium perborate.
a) 10.028 g of borax, 5mL of HCl, and 40mL water used. I got 5.155g of boric acid.
(Na2[B4O5(OH)4]*8H2O(s) + 2HCl(aq)  4H3BO3(s) + 2NaCl(aq) )

b) 20.08g of borax, 4.0488g of sodium hydroxide, 165mL of 3% H2O2, and 40mL water used. I got 29.252g of sodium perborate.
( Na2B4O7*10H2O(aq) + 4H2O2(aq) + 2NaOH(aq) + H2O(l)  2Na2[B2(O2)2(OH)4]*6H2O(s) )

4. Determine the average concentration of each solution titrated.
a) 25mL of 0.1M unstandardized HCl solution, 2drops of phenolphthalein indicator used. Titrated this solution with 15.2mL of standardized sodium hydroxide solution to the stoichiometric end point, a pink color.

[double titration]
b) 10mL of ~0.07M unstandardized Na2B4O7 solution, 50mL of de-ionized water, and 3 drops of methyl red indicator. Titrated with 12.2mL of standardized hydrochloric acid.
 the reaction for the titration of sodium tetraborate with hydrochloric acid solution is: Na2B4O7(aq) + 2HCl(aq) + 5H2O(l)  4H3BO3(aq) + 2NaCl(aq)
c) To the solution remaining from (b), added 5 sccops of sugar mannitol, 2 drops of phenolphthalein indicator. Titrated this solution with 15.1mL of standardized NaOH solution.

5. Determine the percent yield and percent purity of the boric acid and sodium perborate.
a) 0.2 g of boric acid, 5.051g of mannitol, and 50mL of water used. Titrated with 28mL of 0.1M NaOH.
b) 0.25g of Na2[B2(O2)2(OH)4]*6H2O (307.6g/mol) and 30mL of 4M H2SO4 used. Titrated with 7.5 mL of ~0.07M KMnO4.

6. Determine the number of moles and percent by weight of sodium perborate and sodium carbonate in a given sample of Purex.
 sodium carbonate, one of the important industrial chemicals, is the salt of a strong base and a weak acid. Its solution in water is distinctly basic:
CO32-(aq) + H2O(l)  HCO3-(aq) + OH-(aq)
 sodium perborate can also be titrated with acid:
[B2(O2)2(OH)4]2- + 2H+ +2H2O  2H3BO3 + 2H2O2
 the hydrogen peroxide released can subsequently be titrated with a standard solution of potassium permanganate:
5H2O2 + 2MnO4- + 6H+  2Mn2+ + 8H2O + 5O2
a) .921g of Purex, 100mL de-ionized water, 2drops Antifoam C, 5 drops of methyl orange indicator used. Titrated with 8.6 mL of ~0.08M HCl.
b) Added 10 mL of 4M H2SO4 to (a). titrated with 22mL of ~0.015M of KMnO4

##### Solution Summary

The expert examines an the concentration for titration experiments.

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1. Calculate the concentration of the HCl(aq) solution
( 25mL of .1M unstandardized HCl solution, 2 drops of phenolphthalein indicator used. Titrated with 15.2mL of sodium hydroxide solution)

0.2/1000 x 15.2 = 0.003 moles of NaOH
0.003 moles of HCl in 25mL (0.003/25 x 1000 = 0.12M)

2. Calculate the concentration of the Na2B4O7 solution two ways
a) From the HCl titration of the Na2B4O7 solution
( 10mL of ~0.07M unstandardized Na2B4O7 solution, 50mL de-ionized water, and 3 drops of methyl red indicator. Titrated with 12.2mL of standardized hydrochloric acid)

12.2mL of 0.12M = 0.0015 moles in 12.2 mL
2 moles of HCl react with 1 Na2B4O7
0.0015/2 x 10 = 0.0075 moles of Na2B4O7 x 10 = 0.075M

b) From the titration with NaOH
 the reaction for the titration of sodium tetraborate with hydrochloric acid solution is: Na2B4O7(aq) + 2HCl(aq) + 5H2O(l)  ...

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