Precipitate forms when an INSOLUBLE substance is formed. That means you're looking for ions that form insoluble substances when combined with NaOH and LiCl.
That means the ions below are all contenders (use the solubility rules):
Ag+
Hg+
Pb2+
Note: PbCl2 is SLIGHTLY soluble in HOT water, but in standard conditions, PbCl2 is considered as insoluble.
hope this helps.
Sodium hydroxide (NaOH) is an insoluble base that can be used to make copper sulfate. When sodium hydroxide is added to a solution of copper sulfate, a blue precipitate of copper hydroxide forms. This precipitate can be filtered and then reacted with sulfuric acid to produce copper sulfate.
When you raise the pH by adding aqueous NaOH after a precipitate forms in a solution of aqueous sodium benzoate due to a pH decrease, the precipitate likely dissolves. This is because sodium benzoate is the conjugate base of benzoic acid, so at higher pH levels, it remains in solution. The sodium benzoate will revert back to being fully soluble in its aqueous form.
The reaction between C8H5O4K and NaOH will produce potassium salicylate (C7H5KO3) and water. The balanced equation is: C8H5O4K + NaOH → C7H5KO3 + H2O.
The precipitate formed will be calcium carbonate (CaCO3). This is because when ammonium carbonate reacts with calcium nitrate, the insoluble calcium carbonate is formed as a white precipitate, while ammonium nitrate remains in solution.
Reaction between sodium and water produces hydrogen gas. This is not a precipitation, because the gas has a (much) lower density than the water and therefore evolves from it rather than precipitating.
The white precipitate formed when NaOH (aq) is added to MgSO4 (aq) is magnesium hydroxide (Mg(OH)2).
Copper hydroxide is the precipitate.
No, NaOH, sodium hydroxide, is a strong base, not an acid.
Yes, a white precipitate of silver hydroxide (AgOH) will form when solutions of silver nitrate (AgNO3) and sodium hydroxide (NaOH) are mixed. Silver hydroxide is insoluble in water, so it will precipitate out of the solution.
2-methylpropanol does not form a yellow precipitate with iodine in NaOH because it is a tertiary alcohol, which does not undergo the iodine test reaction typically associated with primary and secondary alcohols. The reaction involves the formation of iodoforms, which requires the presence of a hydrogen atom on the carbon adjacent to the hydroxyl group. Since 2-methylpropanol lacks this structure, it does not produce the necessary iodoform compound, resulting in no yellow precipitate.
If acid & base are combined a neutal substance is produced.
The result is a precipitate of Sr(OH)2 as it is insoluble in water. This reaction forms a white precipitate.
When mixed with a solution of sodium hydroxide (NaOH), a precipitate will form with magnesium (Mg) cations. The magnesium hydroxide formed is sparingly soluble in water, leading to the precipitation reaction. Potassium (K) and sodium (Na) cations do not precipitate when mixed with NaOH as their hydroxides are highly soluble in water.
The precipitate is copper(II) hydroxide. The chemical reaction is:2 NaOH + CuSO4 = Cu(OH)2 + Na2SO4
The net ionic equation for NaOH and Na2SO4 when they form a precipitate is simple. It will contain only the atoms that participate in the reaction. Both of these compounds are soluble.
Sodium hydroxide (NaOH) is an insoluble base that can be used to make copper sulfate. When sodium hydroxide is added to a solution of copper sulfate, a blue precipitate of copper hydroxide forms. This precipitate can be filtered and then reacted with sulfuric acid to produce copper sulfate.
The chemical equation is:3 NaOH + FeBr3 = 3 NaBr + Fe(OH)3