Purpose: The following experiment was conducted to prepare standardized solution of sodium hydroxide solution (NaOH) and to determine the concentration of given unknown sulfuric acid (H2SO4) solution.
Analysis: This experiment is divided into two parts. In the first part; the standardized solution of sodium hydroxide is prepared by titrating it with base Potassium hydrogen phthalate (KHP). Phenolphthalein (range 8.3 to 10.0) is used as indicator to determine whether the titration is completed.
Part A: Standardization of a sodium hydroxide solution
NaOH Code sample code: R1
Trial 1
Mass of KHP transferred = 0.41 g
Volume of NaOH used = 21.14 mL Molar mass of KHP = 204.22 g/mol
No. of moles of KHP = Mass of KHP used / Molar mass = 0.41 g
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of moles of KHP = Mass of KHP used / Molar mass = 0.39 g / 204.22 g/mol = 0.001909 moles
Concentration of NaOH = No. of moles / Volume = [0.001909 mol / (21.70 / 1000) L] = 0.0880 M
Trial 5
Mass of KHP transferred = 0.41 g
Volume of NaOH used = 23.59 mL Molar mass of KHP = 204.22 g/mol
No. of moles of KHP = Mass of KHP used / Molar mass = 0.41 g / 204.22 g/mol = 0.002007 moles
Concentration of NaOH = No. of Moles / Volume = [0.002007 mol / (23.59 / 1000) L] = 0.0851 M
Average concentration of NaOH = 0.0949 + 0.0859 + 0.0866 + 0.0880 + 0.0851 / 5
= 0.0881 M
Observations: The Potassium hydrogen phthalate (KHP) was solid and white in color whereas Sodium Hydroxide (NaOH) is a colorless liquid solution. In first trail after adding about 15 to 16 mL of NaOH solution there was repeated appearance of light pink color but would disappear when we swirl the flask. At 21.14 mL of NaOH solution added to KHP and distilled water the pale pink color stays permanent. Same color changes happened in the next four trails when certain NaOH solution reacted with KHP respectively.
Part B: Concentration of Sulfuric Acid solution
H₂SO₄ Sample Code = 16
H2SO4 (aq) + 2NaOH (aq) → 2H2O(l) + 2Na2SO4 (aq)
Trail
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of moles of NaOH = (Average concentration of NaOH) X (Volume of NaOH used)
= 0.0881 M X (37.53/1000) L = 2.3474 moles
No. of moles of H2SO4 = 2.3474/2 = 1.1737 moles
Concentration of H2SO4 = No. of moles / (volume of diluted acid / 1000) L X 10 mL X 1L /1000
= 1.1737 / (25/1000) X (10/1000) L = 0.4694 M
Trail 3:
Volume of diluted acid = 25 mL
Volume of NaOH used = 37.67 mL
Average concentration of NaOH = 0.0881 M
No. of moles of NaOH = (Average concentration of NaOH) X (Volume of NaOH used)
= 0.0881 M X (37.67/1000) L = 2.3387 moles
No. of moles of H2SO4 = 2.3387/2 = 1.1693 moles
Concentration of H2SO4 = No. of moles / (volume of diluted acid / 1000) L X 10 mL X 1L /1000
= 1.1693 / (25/1000) X (10/1000) L = 0.4677 M
Trail 4:
Volume of diluted acid = 25 mL
Volume of NaOH used = 38.32 mL
Average concentration of NaOH = 0.0881 M
No. of moles of NaOH = (Average concentration of NaOH) X (Volume of NaOH used)
= 0.0881 M X (38.32/1000) L = 2.2990 moles
No. of moles of H2SO4 = 2.2990/2 = 1.1495 moles
Concentration of H2SO4 = No. of moles / (volume of diluted acid / 1000) L X 10 mL X 1L /1000
= 1.1495 / (25/1000) X (10/1000) L = 0.4598 M
Trail 5:
Volume of diluted acid = 25
11.) Subtract the mass of the evaporating dish from the mass of the evaporating dish and it's contents. Multiply that number by 10 to get the solubilty in grams per 100 cm3 of water.
10cm3 of 1 molar solution. I will use 3 of each solution to ensure that
Number of moles of maleic anhydride = 0.3g ma* (1 mol ma)/(98.016g ma) = 0.00306mol ma
As more NaOH is added, the pH will become more basic as H2C2O4 .2H2O has been completely neutralized and now an excess of OH- ions are present in the solution.12
We were then to make a base solution of 0.7 M NaOH. In order to standardize
Well, this looks like its using some calculations so what I would do is take my 0.045 M and equal it to the 0.25 mL of NH3 and multiply that by 45.0 mL and multiply it by 10 with an exponent of negative 3. Once all of that is multiplied together we should get an answer of 0.01135 moles of our HCI. Now we can find our “Concentration” Which means we would divide our moles (0.01125) to our vol in liters which is 0.025, once we do that, we get an answer of 0.045M of our NH3. Well, since we are on the topic of pH we know that we can use the formula: pH = -log (H3O+). Then what we would do is plug everything into the formula: pH equals -log (2.4 multiplied by 10 (with an exponent of -5). Once we find the answer to this and we add up all of our calculations, we can come to a conclusion that the answer is: 4.6197 as our pH.
The equation shows how 1 mol of Na2CO3 reacts with 1 mol of H2SO4, so
2. Put the test tube inside a beaker for support. Place the beaker on a balance pan. Set the readings on the balance to zero. Then measure 14.0g of KNO3 into the test tube.
As 10 mL of NaOH was added drop wise, the solution began to have a pink tint. The fuchsia color did not permanently stay until 9mL of NaOH was added. The pH indicator demonstrated when exactly the reaction has neutralized. Because an acid and a base reacted to form water and a salt, the reaction is a neutralization
g. of KI in 10 mL of water. Add the KI solution dropwise to the test
Figure 1: Simple batch homogenous reactor. [Fogler, H. S. (2010, November 22). Essentials of Chemical Reaction Engineering: Mole Balances. Retrieved April 24, 2014, from Pearson Education: http://www.informit.com/articles/article.aspx?p=1652026&seqNum=3]
Compared to the 0.5 M hydrochloric acid that was less concentrated, the more concentrated 2 M hydrochloric acid c...
The purpose of this experiment is to use our knowledge from previous experiments to determine the exact concentration of a 0.1M sodium hydroxide solution by titration (Lab Guide pg.141).