25.0 cm3 of a 0.10 moldm-3 solution of sodium hydroxide was titrated against a solution of hydrochloric acid of unknown concentration. 27.3 cm3 of the acid was required. What was the concentration of the acid? 3. 25 cm3 of a solution of sodium hydroxide reacts with 15 cm3 of 0.1 mol/dm3 HCl.
An aqueous solution of ammonium sulfate is allowed to react with an aqueous solution of lead(II) nitrate. Identify the solid in the balanced equation. A) (NH4 )2 SO4 B) Pb(NO3 )2 C) PbSO4 D) NH4 NO3 E) There is no solid formed when the two solutions are mixed. ____ 11. An aqueous solution of sodium carbonate is reacted with an aqueous solution of calcium chloride.
For example, the reaction |BaCl2•2H2O ( BaCl2 + 2H2O |(2) | is reversible, and if water is added to the anhydrous salt BaCl2, formation of BaCl2•2H2O takes place: |BaCl2 + 2H2O ( BaCl2•2H2O |(3) | The reaction of dehydration of hydrated ferrous sulfate |FeSO4•7H2O ( FeSO4 + 7H2O |(4)
The chemical reaction used to find this constant is as follows: MgC2O4 (s) ↔Mg(aq)2++ C2O4 (aq)2- Kc= Mg2+[C2O42-][MgC2O4] Ksp=Mg2+[C2O42-] The solid salt magnesium oxalate is prepared through the following precipitation reaction: Mg(SO4)(aq)+NaC2O4 (aq) → MgC2O4 (s)+NaSO4 (aq) Next, the concentration of the Mg2+ and C2O42- ions is found through a redox titration. This redox titration uses a standardized potassium permanganate solution. The potassium permanganate solution is standardized by titrating it with samples of iron(II)ammonium sulfate hexahydrate . The end point is reached when the solution has turned light purple which is a result of excess amounts of MNO4-. This reaction can be summed up using the following formula: 5Fe2++8H++MnO4- →5Fe3++Mn2++4H2O After standardization, the potassium permanganate solution is then titrated with 3 different magnesium oxalate solutions.
Micro – scale Reduction of a Ketone to an Alcohol: Benzophenone to Diphenylmethanol with Sodium Borohydride Abstract: Purpose of experiment 2.1 was to perform reduction reaction of benzophenone to diphenylmethanol with sodium borohydride as a reducing agent. The reducing agent was used in excess to ensure complete reduction of the carbonyl group. The product was isolated as a solid by filtration and its purity was checked using Thin Layer Chromatography with different ratios of mixture of polar and non – polar solvents and by checking its melting point, which was 520C - 620C. Infra – Red spectroscopy was not performed, however previously printed graphs were compared and analysed based on tables in Chemistry Laboratory Manual. Purpose of the Experiment: To produce diphenylmethanol from the reduction of benzophenone by using sodium borohydride as a reducing agent.
How could it have been improved? (b) Based on your evidence, which anion could be used to precipitate most of the metal cations? (c) Which anion could be used to selectively remove silver ions from solution? Why? (d) What evidence suggests that nitrate compounds are soluble in water?
Contents Abstract – Page 2 Summary of Results – Page 3 Focus Question – Page 5,6 What if… - Page 6 Confidence Report – Page 7,8 Abstract Summary of Results Observations of Reactions – Trials 1 , 2 & 3 | Dissolving Barium Iodide and Zinc Sulfate in deionized water | As the substances dissolve the water becomes cloudy and acquired a white tint. | Centrifuging up the solution obtained in previous reaction | The precipitate and the liquid seperated, and there was more precipitate in one of the test tubes. The precipitate was a thick white color, not transparent. | Heating the Precipitate in a boiling tube containing 2 boiling chips | The mixture began producing a large amount of bubbles, as we continued heating it only white powder remained. | Observations of Chemicals | Zinc Sulfate | Powder of a white solid | Barium Iodide | Powder of a white solid.
3 x (C H5 N) = C3H15N3 Hydrated compounds Solving process: 1st- the difference between the initial mass and that of the dry sample is the mass of water that was driven off. Mass of hydrate minus mass of dry sample equals the mass of water 10.407 – 9.520 = 0.887 g 2nd- The mass of dry BaI2 and the mass of water are converted to MOLES. 9.520 g BaI2 x 1 mol BaI2 ∕ 391 g BaI2 = 0.0243 mol BaI2 anhydrate 0.887 g H2O x 1 mol H2O / 18.0 g H2O = o.o493 mol H2O 3rd: Dividing both results by the amt of 0.0243 mol, we get a ratio of 1 to 2.03, or 1 to 2, since the formula must have full numerical integers of water molecules, in other words no fractions of a water molecule. Thus, for every 1 mole of BaI2, there are two moles of water. The formula for the hydrate is written as BaI2 • 2H2O And it is named barium iodide dihydrate.
The reaction that occurred with this step was displacement and metathesis in the form of gas formation. The balanced equation of this step looks as follows: CuSO4aq+Zns→Cus+ZnSO4(aq) Once this step was finished, the remaining copper was retrieved. First, to recover the copper HCl was added to remove all the zinc. When this happened, a yellow tint was observed in the liquid, as well as bubbling as the zinc was broken down. Once the copper dried out, it was weighed and came to a total of 240 mg.
Results : Yield (mass in gram) = __________________0.128 g_____________________ Melting point of pure phenyl benzoate (OC) = ___________70O______________ Melting point of your phenyl benzoate (OC) = ___________69O______________ Questions : 1. Why do we use NaOH solution to dissolve phenol rather than water in step 1 ? Because NaOH can be use to change the phenol to ion form, which can be more soluble in the aqueous