What occurs when (i) sodium hydroxide is added to silver nitrate; and (ii) ammonia solution is added to the resultant precipitate?

Answer 1

You know that solubility rules follow a hierarchy, and that complex formation can occur with certain ligands. Here, (a) addition of #NH_3# will give rise to a colourless solution; and (b) addition of #HCl# will give rise to a precipitate.

Since all nitrates are soluble and all hydroxides—aside from those of the alkali metals—are insoluble, the first reaction can be represented as follows:

#AgNO_3(aq) + NaOH(aq) rarr Ag(OH)(s)darr + NaNO_3(aq)#
The down arrow signifies that the silver crashes out of solution as an hydroxide. The "silver hydroxide" is very poorly characterized. Addition of (a) #NH_3# will form a complex ion with silver, which we may represent as:
#Ag(OH)(s) + 2NH_3(aq) rarr# #Ag(NH_3)_2^+#(aq) + #OH^-#(aq)
Addition of #HCl(aq)# will result in an acid base reaction with #Ag(OH)#. The silver salt will go up momentarily, however, #AgCl#, a very insoluble species will crash out as a (momentarily) white solid, i.e.
#Ag(OH)(s) + HCl(aq) rarr AgCl(s)darr + H_2O#
On standing you will see the white solid darken as silver is reduced to form finely divided silver metal. This is more of an undergraduate than an A level treatment. If you're interested, what I have represented as #Ag(OH)# is probably a hydrous oxide of silver, i.e. #Ag_2O*H_2O#. No one really knows, because it is so insoluble. You will notice that the white precipitate darkens on standing as metallic silver precipitates.

How can I remember all of this? Perform the reaction in the lab, pay close attention to what happens, and ask your professor to confirm what is anticipated.

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Answer 2

(i) When sodium hydroxide is added to silver nitrate, a white precipitate of silver hydroxide is formed according to the following equation:

[ AgNO_3 (aq) + NaOH (aq) \rightarrow AgOH (s) + NaNO_3 (aq) ]

(ii) When ammonia solution is added to the resultant precipitate (silver hydroxide), it dissolves to form a colorless, complex ion known as diamminesilver(I) hydroxide, according to the following equation:

[ AgOH (s) + 2NH_3 (aq) \rightarrow [Ag(NH_3)_2]OH (aq) ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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