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
this is the best volume to use as it is about ¾ of a test tube full,
10cm3 of 1 molar solution. I will use 3 of each solution to ensure that
We were then to make a base solution of 0.7 M NaOH. In order to standardize
For part C, the concentration of was determined to be 1.01 mol/L, 0.973 mol/L, and 1.158 mol/L. These results show a relatively closed to the accepted 1.00mol/L of NaOH. The differences of these results are understandable since the concentration of NaOH would changes over time because during the transfer of NaOH powder in part A, it was exposed to the air, thus it could reacts with CO2 in the atmosphere to produce Na2CO3 and water, therefore, changing the concentration of NaOH. Furthermore, the NaOH could also react with the glass thus it wills also reducing its concentration. However, all of the concentration of NaOH that was determine are maximum of 0.158mol/L differences compare to the standard 1.00 mol/L, therefore, it can be concluded that the result are accurate.
The first step that I did to find the number of grams water can be produced when 11.7 moles of ethane that reacts with the excess oxygen gas was to balance the equation like this 2 C2H6 + 7 O2 ----> 4 CO2 + 6 H20. The second step that I did to find the number of grams water can be produced when 11.7 moles of ethane that reacts with the excess oxygen gas was to find the mole ratio which is 2 moles of C2H6 : 6 moles of H20. The third step that I did to find the number of grams water can be produced when 11.7 moles of ethane that reacts with the excess oxygen gas was to multiply this by the mole ratio like this 6 moles of water / 2 moles of ethane * 11.7 moles of ethane = 35.1 moles of water. The fourth step that I did to find the number of grams
The equation shows how 1 mol of Na2CO3 reacts with 1 mol of H2SO4, so
In my experiment, I will use an overall volume of 50 cm³ of 2moles of
-2152.7 x (56.1 / 1.37) = -88150.7 J.mol. 1. H = -88.15 kJ.mol. Hess' law states that: 1"The total enthalpy change for a chemical reaction is independent of the route by which the reaction takes place, provided initial and final conditions are the same.
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.
= = = Concentration of sucrose solution (M) Length of potato (cm) Before After Difference 0.0 5.
Sodium Thiosulphate's Effect on the Rate of Reaction with Hydrochloric Acid Aim. I am going to investigate how varying the concentration of sodium Thiosulphate affects the rate of reaction with Hydrochloric Acid. Prediction: The equation for the reaction is: [ IMAGE] Sodium Thiosulphate + Hydrochloric Acid Sodium Chloride + Water +. Sulphur + Sulphur dioxide [ IMAGE] Or: Na2S2O3 + 2HCl S + 2NaCl + H2O + SO2.
Moles Volume HCl Volume Water 2 M 10 cm 3 0 cm 3 1.5 M 7.5 cm 3 2.5 cm 3 1 M 5 cm 3 5 cm 3 0.5 M 2.5 cm 3 7.5 cm 3
Electrolysis Investigation Planning In this investigation, I will assess how changing the electric current in the electrolysis of acidified water affects the rate at which hydrogen gas is produced. The solution to be electrolysed is made up using acid and water. It is of little consequence what acid is used however in this case I will use Sulphuric acid (H2SO4). When H2SO4 is put in water it is dissociated and forms ions: H2SO4 → 2H (2+) + SO4 (2-) Ions are also present from the water in the solution: H2O → H (+) + OH (-) During the electrolysis process, the positive hydrogen ions move towards the cathode and the negative hydroxide and sulphate ions move towards the anode.
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
In this experiment the Sodium Hydroxide solution went through three different phases where its quality and quantity changed. The first phase was called I. Preparing Approximately 0.1M NaOH, 1000mL of clear distilled water was boiled and then chilled to room temp.