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Copper can be obtained from copper sulfate solution by electroplating it onto an electrode or by adding a metal higher in the electromotive series than copper, such as iron, to the solution. The more active metal will dissolve by displacing copper in metallic form from the copper sulfate.
The easiest way of making sodium sulfate solution is by adding sulfuric acid to sodium hydroxide. If the pure material is required, then evaporate the water off.
chemical
A concentrated solution can be weakened by adding more of the solution material (usually water) to dilute it.
Adding an acid to a solution will lower it's Ph, but adding a base will raise it's Ph.
The "excess" metallic copper produced by adding zinc metal to a copper sulfate solution comes from exchanging zinc atoms from the metal for copper atoms from the copper sulfate solution. During the reaction, the zinc atoms are ionized to cations and the copper cations from the solution are reduced to neutral atoms.
Sulfate ions in a solution are verified by adding BaCl2. If an acid was not added, it might be confused with BaCO3, if the solution has carbonate ions.
Because many chemical procedures need dried reagents
Copper can be obtained from copper sulfate solution by electroplating it onto an electrode or by adding a metal higher in the electromotive series than copper, such as iron, to the solution. The more active metal will dissolve by displacing copper in metallic form from the copper sulfate.
Anhydrous simply means 'without water'. The opposite is a hydrated substance, in which salt crystals contain water as an integral part of the crystal. For example, anhydrous cobalt(II) chloride, CoCl2, is a blue powder. Add water and you form cobalt(II) chloride hexahydrate, CoCl2.6H2O, which is a pink colour. Anhydrous cobalt chloride can be used to test for the presence of water because of this dramatic colour change. In addition, hydrated copper(II) sulfate, CuSO4.5H2O, forms bright blue crystals. Heating them drives off the water of crystallisation and anhydrous CuSO4, a grey/white powder, is formed. Adding water will reform a blue solution of the hydrated salt.
To identify the presence of sulfate ions in a solution, follow these steps: To acidify the sample, add a few drops of dilute hydrochloric acid (HCl) to the sample. Then, add a few drops of dilute barium chloride (BaCl2) solution to the sample. If sulfate ions are present in the solution, a white precipitate of barium sulfate (BaSO4) will form. The reaction is: Ba²⁺ (aq) + SO4²⁻ (aq) → BaSO4 (s) For example, if we have a solution of magnesium sulfate (MgSO4), the reaction would be: BaCl2 (aq) + MgSO4 (aq) → BaSO4 (s) + MgCl2 (aq) By observing the formation of the white precipitate, we can confirm the presence of sulfate ions in the sample.
Sodium sulfate is highly soluble in water, but insoluble in most organic solvents. If you want to increase its solubility in water (as for any salt), you can heat the solution or remove one of the products (sodium ions or sulfate ions) from solution. I can't think of any insoluble sodium salts, but barium sulfate (BaSO4) is insoluble in water. Thus, adding barium chloride (or some other soluble barium salt) will remove sulfate from the equilibrium (due to BaSO4 precipitation) and increase the solubility of sodium sulfate.
The easiest way of making sodium sulfate solution is by adding sulfuric acid to sodium hydroxide. If the pure material is required, then evaporate the water off.
Adding a base
chemical
A concentrated solution can be weakened by adding more of the solution material (usually water) to dilute it.
Adding an acid to a solution will lower it's Ph, but adding a base will raise it's Ph.