Introduction: The unknown acid molarity will be determining by titration method. Titration is a process depends on concentration of known solution to another solution until the solute in the another solution completely react. Standard solution is the solution of known concentration that used in titration. In this experiment, NaOH was the titrant (base) however, the two analyte which used were HCl and H2SO4. The chemical reaction equations are molecular and ionic molecular equation for (NaOH) and (HCL) is: HCl (aq) + NaOH (aq) --> H2O (l) + NaCl (aq) The Net Ionic equation is: H+ (aq) + OH- (aq) --> H2O (l) Molecular equation for (NaOH) and (H2SO4) is: H2SO4(aq) +2NaOH(aq) --> Na2SO4+ 2H2O The Net Ionic equation is: H+ (aq) + OH- (aq) --> …show more content…
It is highly reactive and it is solutions are not considered to be carcinogens, the molar mass for H2SO4 is 98.09 g/mol. However, hydrochloric acid is corrosive and it is colorless, with the molecular formula HCL. The molar mass for HCl is 36.46 g/mol. Phenolphthalein is flammable and harmful; it is absorbed through the skin. To be safe avoid contact with eyes, skin, and clothing. In case of skin contact, eye contact, ingestion or inhalation, lab coats, gloves and safety spectacles were worn at all times throughout the experiment. In this experiment the alkali NaoH(aq) has been collected about 100 cm3 in a large beaker and about 75 cm3 were collected for the acid in a small …show more content…
Step 4: Calculate the molarity of the analyte HCl(aq) M(HCl) = n/V = 2.235x10^-3 / 0.0250 = 0.0894 M Finding with H2SO4: Find the Molarity of H2SO4(aq) with analyte NaOH(aq). Step 1: Alkali NaOH(aq) (ANALYTE) Acid H2SO4 (aq) (TITRANT) V(NaOH) = 31.50 cm3 - 0.0315 dm3 V(H2SO4) = 25 cm3 - 0.0250 dm3 M(NaOH) = 0.100 M M(H2SO4) =? Step 2: M = n/V, n(NaOH) = M x V = 0.100 x 0.0315 = 3.15x10^-3 Step 3: The chemical equation: H2SO4(aq) +2NaOH(aq) --> Na2SO4+ 2H2O 1mol 2mole Molar ratio = H2SO4:NaOH = 1:2 = ? : 3.15x10^-3 Thus, n(H2SO4)= 3.15x10^-3 /2 = 1.1175 x10^-3 mol This mean that 1.1175x10^-3 mole of H2SO4(aq) is used in this experiment. Step 4: Calculate the molarity of the analyte HCl(aq) M(H2SO4) = n/V = 1.1175 x10^-3 / 0.0250 = 0.0447
A two-hundred and fifty milliliter(ml) flask was zeroed out on the scale. Fifty milliliters of HCl was put into the flask and weighed. The unknown mixture was poured into the HCl while
Can Alka-Seltzer Act as a Buffer Against Acid Rain? Background: Acid rain is a product of rainfall being tainted by atmospheric pollution. This also means that it has high levels of hydrogen ions. When acid rain falls it has negative effects on natural environments such as aquatic life, plants, and infrastructure.
A 0.2057 M stock solution of NaOH was added to the plastic syringe while the acidic solution was being made. 1.329 g of unknown acid P was weighed on the electronic scale and recorded. The unknown solid acid was placed into a 100 mL volumetric flask and filled to the line with a 0.1 M stock solution of KCl. The solution was agitated to aid in the dissolving of the solid acid solution. When the acid was completely dissolved 25 mL of distilled water was added to a clean 100 mL beaker, along with 25.00 mL of acidic solution using a volumetric pipette.
To do this we took our answers from the other equations and subtracted them. The two equations for the two antacids are: We then calculated the molarity of the mixture before titration using one of our answers from before. The two equations for the Tums and Up and Up are: We then calculated the percent error for each antacid using the number of moles expected to be neutralized, theoretical yield, and the number of moles actually neutralized, experimental value.
(28.54g-24.21g). we found the mass of the anhydrous salt to be 2.57g by subtracting the mass of the crucible + salt by the mass of the crucible after heating, (26.78g-24.21g). we found the moles of anhydrous salt to be .016 by dividing the mass of the anhydrous salt by the molar mass of the anhydrous salt, (2.57g/158.53 g/mol). To find the mass of H20 driven off we subtracted the mass of hydrated salt by the mass of anhydrous salt, (4.33g-2.57g=1.76g) Then we calculated the moles of H20 driven off to be .098mol by dividing the mass of H2O driven off by 18, (1.76g/18).
There is one mole of OH- in the solution since NaOH goes to Na+ and OH-. Trial 1: 25.65mL NaOH x 0.100mol/1000mL = 2.57 x 10-3 mol NaOH = 2.57 x 10-3 mol HA = 2.57 x 10-3 mol H+. The equivalent mass is 0.356g Acid / 2.57 x 10-3 mol H+ = 139g/mol H+ Trial 2: 49.57mL NaOH x 0.100 mol / 1000 mL = 4.96 x 10-3 mol NaOH = 4.96 x 10-3 mol HA = 4.96 x 10-3 mol H+. The equivalent mass is 0.644g Acid / 4.96 x 10-3 mol H+ = 130.g/mol H+ Average =
Finding the empirical formula for hydrated copper sulfate using calculations to find the amount of each element present in the copper ion, sulfate ion, and water while also comparing the empirical formula to a literature value. Christian Cooper Alexis Powers CHM1210-18M/Gregory Bowers 11-5-15 Purpose: To begin, there are several different goals, techniques, and claims to note in the experiment involving hydrated copper sulfate. The overall goal of this experiment is to find the empirical formula and compare it with a literature value. Yet, in finding the empirical formula of hydrated copper sulfate, there are several process for it to get through, like finding the percentages of copper, water, and finally sulfate.
The Freiburg Study Melaleuca’s Peak Performance Pack was tested in what was called “The Freiburg Study” where 48 healthy human subjects of various ages were given the product. Twenty four of the subjects had metabolic syndrome- with blood pressure, cholesterol, or body weight that was elevated but all still within normal range. The other 24 subjects had healthy markers closer to what would be considered ideal. They were tested for free radical activity, cholesterol levels, blood lipid levels, C-reactive protein (determines heart health), and glucose levels, insulin response and inflammation, heart rate and blood pressure (Freiburg Study).
Verna Wang Hannah Palmer CHEM 101-069 Lab 11-19-16 Stoichiometry and Limiting Reagents Lab Report Purpose: We are using the reaction of sodium hydroxide and calcium chloride to illustrate stoichiometry by demonstrating proportions needed to cause a reaction to take place. Background: Just like a recipe would call for a specific amount of one ingredient to a specific amount of another, stoichiometry is the same exact method for calculating moles in a chemical reaction. Sometimes, we may not have enough of or too much of one ingredient , which would be defined as limiting and excess reagent, respectively.
The reagents used were Diphenylamine reagent which contains concentrated H2SO4. The standard solution used for this test is the deoxyribose standard solution. In the sample, only a faint blue solution appeared, which indicates a small presence of deoxyribose. In test for Phosphate, the standard solution was the Phosphate solution and the reagents used were concentrated H2SO4, concentrated HNO3, 2.5% ammonium molybdate solution.
IV. Data and observations Mass of beaker (g) 174.01 Mass of beaker + NaOH pellets (g) 174.54 Mass of NaOH pellets 0.53 TRIAL 1 TRIAL 2 Mass of potassium acid phtalate (KHP) (g) 0.15 0.15 final buret reading (ml) 30.75
Practical I: Acid-base equilibrium & pH of solutions Aims/Objectives: 1. To determine the pH range where the indicator changes colour. 2. To identify the suitable indicators for different titrations. 3.
Enthalpy of neutralization The purpose of this experiment is to determine the enthalpy change for the reaction between aqueous sodium hydroxide (NaOH) and aqueous hydrochloric acid (HCl). Introduction A neutralization reaction is a chemical reaction where a base and an acid react with each other.
Introduction The goal of the experiment is to examine how the rate of reaction between Hydrochloric acid and Sodium thiosulphate is affected by altering the concentrations. The concentration of Sodium thiosulfate will be altered by adding deionised water and decreasing the amount of Sodium thiosulphate. Once the Sodium thiosulphate has been tested several times. The effect of concentration on the rate of reaction can be examined in this experiment.
That caused a new initial reading of NaOH on the burette (see Table1 & 2). The drops were caused because the burette was not tightened enough at the bottom to avoid it from being hard to release the basic solution for titrating the acid. The volume of the acid used for each titration was 25ml. The volume of the solution was then calculated by subtracting the initial volume from the final volume. We then calculated the average volume at each temperature.