Calculate the concentration of grams of sodium stearate per milliliter of diluted solution. To do this, multiply the concentration of sodium stearate in the dishwashing liquid by the dilution of the solution (1.50 mL dishwashing liquid per 100 mL solution). Answer = 1.5 *10^-4 g/mL 4. Calculate the number of moles of sodium stearate in a single layer. To do this, first take the number of drops used to achieve the monolayer (1 drop) and convert it to mL using the calibrated number of drops per mL.
Repeat the titration until there are two titres within 0.1cm3 of each other. Record results in a suitable table. Results: Rough Titre: 7.653 First Run: 6.553 Second Run: 6.453 Third Run: 6.553 Calculations: During the titration, iron(II) ions are oxidised to iron(III) ions and manganate(VII) ions are reduced to manganese(II) ions. The equation is as follows: 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) ? 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) The above equation shows that one mole of manganate(VII) ions reacts with 5 moles of iron(II) ions in acid solution.
White precipitate shows the presence of chloride (Cl-). Chloride anion equation: HCl(aq) + AgNO3 (aq) → HNO3 (aq) + AgCl(s). The nitrate anion test involves cooling a mixture containing 1 mL of test solution and 3mL 18M H2SO4. 2mL is poured down the inner test tube side and the presence of a brown ring shows nitrate (NO3-) to be present. The carbonate anion test mixes 1 mL of test solution and drops of 6M HCl.
For the third part of the experiment, the process of part 1 was repeated using sodium hydroxide. 5 drops of 0.10 M sodium hydroxide were placed in wells 1 and 2, and 5 drops of distilled water were placed in wells 2 through 6. A toothpick was used to stir the mixture in well 2, resulting in 0.05 M of sodium hydroxide. A pipet was used to extract the solution from well 2 and 5 drops were added to well 2. The dilution process was repeated for the remaining wells, as completed in part 1.
C. The final pH is 7. D. Water is produced. 2. Which salt could be prepared by a method involving crystallization as a final stage? A. barium sulphate B. calcium carbonate C. silver chloride D. sodium nitrate 3.
This precipitate must be a combination of the K + or Pb 2+ cation and the CrO4 2- or NO3 2- anion. Since most potassium salts and nitrates are soluble in water, PbCrO4 must be the precipitate. The equation for this reaction is: Pb2+ (aq) + CrO4 2- (aq) PbCrO4 (s) Equipment: * 100 mL beaker * 250 mL Erlenmeyer flask * Buchner funnel * Stirring rod * Measuring cylinder * Test tube * Electronic balance Material: * 1.0 M Potassium chromate solution * 1.0 M Lead (II) nitrate solution 3. Procedure Step 1 50mL of 1.0 M Potassium Chromate solution is added in a 100 mL beaker. Step 2 50mL of 1.0 M Lead(II) Nitrate
Reactions with Acids and Alkalis Introduction: The purpose of this experiment is to perform five different experiments with Acids and Alkalis. [Hydrochloric acid (HCl), Sodium hydroxide (NaOH)] by finding pH values. Exp 1: The purpose of this experiment is to determine the pH value by adding Acid and alkaline solution in water with Universal indicator. Exp 2 The purpose of this experiment is to define the reaction that happens to the magnesium (Mg) when it is added to HCl acid and NaOH alkali. Exp 3 The purpose of this experiment is to determine the reaction that happens to the calcium carbonate when it is added to HCl acid.
LAB 1 NAME: Amy Rampersad. DATE: Saturday 7th April, 2012. TITLE: PREPARATION OF A SOLUBLE SALT BY A TITRATION METHOD. AIM: To prepare Sodium Chloride crystals by Titration of Sodium Hydroxide with Hydrochloric Acid. APPARATUS: Burette (50 cmᶾ), Pipette (25 cmᶾ), two conical flasks (250 cmᶾ), two beakers (250 cmᶾ), funnel, wash bottle, retort stand, boss and clamp, evaporating dish, pipette filler, hot-plate.
What is the percent yield of the reaction? [Ans. : 43.21] 3) A 5.6780 gram sample of the compound Na2XO3 was dissolved in water then reacted with excess calcium chloride: Na2XO3(aq) + CaCl2(aq) →CaXO3(s) + 2NaCl(aq) A total of 5.4842 grams of CaXO3 were collected. What is the identity of the element “X?” [Ans. : Se, calculated molar mass 78.93 g] 4) A 0.3528 gram mixture of H2SO4 (molar mass 98 grams) and H3PO4 (molar mass 98 grams) was mixed with a little water and titrated with 9.70 mL of 1.000 M NaOH: 3NaOH(aq) + H3PO4(aq) →3H2O(l) + Na3PO4(aq) 2NaOH(aq) + H2SO4(aq) →2H2O(l) + Na2SO4(aq) What was the percent by mass H2SO4 in the original mixture?
TITRATION OF AN ACID (A PREPARED STANDARD SOLUTION OF KH₅O₄C₈) AGAINST A BASE (NAOH) USING PHENOLPHTHALEIN AS AN INDICATOR BY GRACE The aim of this experiment is to prepare a standard solution of potassium hydrogen phthalate (KH₅O₄C₈) and then use it to calculate the concentration of sodium hydroxide by titrating the acid (KH₅O₄C₈) against the base (NaOH). Before the whole experiment could take place, some apparatus were needed which included the following; A weighing balance, burette, pipette, a conical flask, clamp and then the setup was as below; THEORY To calculate for the moles of KH₅O₄C₈, I used n (mol) =m (g)/M (gmol⁻ⁱ (JOHN GREEN AND SADRU DAMJI, PG 6 OF CHAPTER 1.THIRD EDITION). Whereby m=mass of the acid, M=molar mass of the acid and n=number of moles. Molar mass of KH₅O₄C₈ is 204.1g/mol and its mass is 1g Therefore=1g/204.1gmol⁻ⁱ n=0.0048996mol Further more, to calculate for the concentration of the acid, I used C(moldm⁻3.) =n (mol)/v (dm3) Whereby c=concentration, n=number of moles and v=volume used.