Essay On Standardization Of Sodium Hydroxide

1424 Words3 Pages

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 …show more content…

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 …show more content…

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

Open Document