When excess of hydrogen or hydronium ions are present in the solution , the solution will have pH less than 0. Phenolphthalein possesses dark orange color to the solution having pH less than 0. Hence, phenolphthalein will possess dark orange color in the presence of excess of hydrogen or hydronium ions.
In acidic conditions (ie. excess H+ ions) it is colourless. It turns to a bright pink in basic conditions. (alkali, OH-)
it's colorless
The color is pink.
pink
Copper metal itself does not react with sodium hydroxide. But when NaOH is added to a solution of copper ions, it would form a light blue precipitate, which is copper(II) hydroxide, and will NOT dissolve with the excess alkali.
Al(OH)3 (s) + OH- (aq) -------> [Al(OH)4]- (aq) Aluminium hydroxide undergoes further reaction with hydroxide ion to from a complex ion, which is of course, soluble in water. Some other metal hydroxides also have similar reaction, like zinc hydroxide, and lead (II) hydroxide.
Maybe if you asked the question properly, we could answer.
Zinc hydroxide Zn(OH)2 is an inorganic chemical compound. It also occurs naturally as 3 rare minerals: wülfingite (orthorhombic), ashoverite and sweetite (both tetragonal).Like the hydroxides of other metals, such as lead, aluminium, beryllium, tin and chromium, zinc hydroxide (and zinc oxide), is amphoteric. Thus it will dissolve readily in a dilute solution of a strong acid, such as HCl, and also in a solution of an alkali such as sodium hydroxide.It can be prepared by adding sodium hydroxide solution, but not in excess, to a solution of any zinc salt. A white precipitate will be seen: Zn2+ + 2OH- → Zn(OH)2.If excess sodium hydroxide is added, the precipitate of zinc hydroxide will dissolve, forming a colorless solution of zincate ion: Zn(OH)2 + 2OH- → Zn(OH)42-. This property can be used as a test for zinc ions in solution, but it is not exclusive, since aluminum and lead compounds behave in a very similar manner. Unlike the hydroxides of aluminum and lead, zinc hydroxide also dissolves in aqueous ammonia to form a colourless, water-soluble ammine complexThe reason that the zinc hydroxide will dissolve is because the ion is normally surrounded by water ligands; when excess sodium hydroxide is added to the solution the hydroxide ions will reduce the complex to a -2 charge and make it soluble. When excess ammonia is added, it sets up an equilibirum which provides hydroxide ions; the formation of hydroxide ions causes a similar reaction as sodium hydroxide and creates a +2 charged complex with a co-ordination number of 4 with the ammonia ligands - this makes the complex soluble so that it dissolves
This reaction is to be carried out with Sodium hydroxide dissolved in water and Zinc oxide. The product is Sodium Tetrahydroxidozincate(2-) or Simply Sodium Zincate.ZnO + 2 NaOH + H2O ----> Na2Zn(OH)4
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The reaction of zinc nitrate and excess sodium hydroxide begins with precipitation of zinc hydroxide ( Zn(OH)2 ), followed by dissolvement after adding excess sodium hydroxide ( 2 OH- ) to formation of zinc aat-ions ( [Zn(OH)4]2- )
There is no reaction , because of the Common Ion Effect. The Common Ion is the Hydroxide.
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presence of urea in? excess presence always leads to a defect.
One such salt would be aluminum chloride since it is soluble but when reacted with ammonium hydroxide, the insoluble aluminum hydroxide forms a precipitate. Not sure what is meant by "is insoluble in excess", however.
precipitated Fe(OH)3 = ferric hydroxide or Fe(OH)4-(aq) = ferrate anion in solution, when in excess of hydroxide
aluminum hydroxide magnesium hydroxide simethicon
PH increase
Ammonium hydroxide dissolves anything that is less strong than itself. The white precipitate of zinc hydroxide is not the whole component. Therefore, it is not as strong.
Put drops of Sodium, Potassium, or Ammonium Hydroxide in it. The Cupric hydroxide will precipitate out in blue colour. Dont put excess hydroxide or there will be a formation of another intense blue complex compound
Aluminium hydroxide = Al(OH)3 , and Zinc hydroxide = Zn(OH)2 , is redissolved in excess Sodium hydroxide = NaOH (in water solution) : Al(OH)3 + NaOH + H2O ----> NaAl(OH)4 Zn(OH)2 + NaOH + H2O ----> Na2Zn(OH)4