By adding sulfate ions (SO4^2-) to a solution containing Pb^2+ and Ba^2+, lead sulfate (PbSO4) will precipitate due to its low solubility product. Barium sulfate (BaSO4) will also precipitate, as it is also insoluble in water.
When sodium hydroxide is added to lead, a white precipitate of lead(II) hydroxide forms. This reaction can be represented by the chemical equation: Pb(NO3)2 + 2NaOH → Pb(OH)2 + 2NaNO3. Lead(II) hydroxide is insoluble in water and will precipitate out of the solution.
To precipitate magnesium ion, you can add a precipitating agent such as sodium hydroxide (NaOH) to the magnesium nitrate solution. This will cause magnesium hydroxide (Mg(OH)2) to form as a precipitate. You can then filter the solution to separate the precipitate from the liquid.
When iodide is added to silver nitrate, a chemical reaction occurs, resulting in the formation of silver iodide precipitate. This can be represented by the equation: AgNO3 + KI -> AgI(s) + KNO3. The silver iodide formed is insoluble in water and appears as a yellow precipitate.
Test: Add aqueus sodium carbonate solution, then warm it. Observation: Formation of white precipitate. Deduction: The presence of aluminum ions, Al3+ Or Test: Observation: Add aqueous sodium hydroxide White precipitate occur Add aqueous sodium hydroxide (excess) & warm White precipitate solute Test with red litmus paper Changes red - blue Deduction: The presence of aluminum ions, Al3+
yes it will precipitate DNA if your lysing nuclei; add benzamidine hydrochloride though as a protease inhibitor.
When sodium hydroxide is added to lead, a white precipitate of lead(II) hydroxide forms. This reaction can be represented by the chemical equation: Pb(NO3)2 + 2NaOH → Pb(OH)2 + 2NaNO3. Lead(II) hydroxide is insoluble in water and will precipitate out of the solution.
The solid formed when adding lead nitrate and potassium iodide and filtering the mixture is lead iodide. Lead iodide is a yellow precipitate that forms by the reaction between lead nitrate and potassium iodide.
Take a few drops of both samples and add some lead nitrate. A yellow precipitate indicates lead iodide and it gives the inference that it contains iodide ions, hence the solution of sodium iodide.
One method to separate lead chloride from a mixture of lead chloride and silver chloride is to dissolve the mixture in water, then add hydrochloric acid to precipitate the lead chloride while keeping the silver chloride in solution. The precipitated lead chloride can then be filtered out. Another method is to use selective precipitation by adding a potassium chromate solution, which will form a yellow precipitate with the lead chloride while leaving the silver chloride in solution.
Add the compound to nitric acid untill the solution is near colourless, then following this add 2-3 drops of silver nitrate, if a white precipitate forms then the chloride ion is present. to identify if the Cation is lead, try doing a flame test with an emission spectra. cl- + ag+ ---> AgCl
One way to differentiate between lead nitrate and silver nitrate is to add a few drops of dilute hydrochloric acid to each compound. Lead nitrate will produce a white precipitate of lead chloride, while silver nitrate will produce a white precipitate of silver chloride. Another method is to observe the color of the compounds - silver nitrate is white, while lead nitrate is colorless.
Usually, what you're doing is to add a reagent (acid or base, for example) so that the precipitate redissolves. That allows you to run further tests.
To precipitate magnesium ion, you can add a precipitating agent such as sodium hydroxide (NaOH) to the magnesium nitrate solution. This will cause magnesium hydroxide (Mg(OH)2) to form as a precipitate. You can then filter the solution to separate the precipitate from the liquid.
When iodide is added to silver nitrate, a chemical reaction occurs, resulting in the formation of silver iodide precipitate. This can be represented by the equation: AgNO3 + KI -> AgI(s) + KNO3. The silver iodide formed is insoluble in water and appears as a yellow precipitate.
Add silver nitrate solution. flouride is precipitated as silver fluoride
Test: Add aqueus sodium carbonate solution, then warm it. Observation: Formation of white precipitate. Deduction: The presence of aluminum ions, Al3+ Or Test: Observation: Add aqueous sodium hydroxide White precipitate occur Add aqueous sodium hydroxide (excess) & warm White precipitate solute Test with red litmus paper Changes red - blue Deduction: The presence of aluminum ions, Al3+
Add vinegar and measure the protein precipitate. Mass of precipitate/mass of shake x100= % protein