Which of the following solutions will not form a precipitate when added to 0.10M NaOH?


a)0.10 M MgBr2
b) 0.10 M NH4Br
c) 0.10 M FeCL2

Well, if we want to find a solution that won't form a precipitate, we need to think about the solubility rules. NaOH is a strong base, so we're looking for compounds that are soluble in water.

Let's break this down:

a) 0.10 M MgBr2 - Sorry, but this solution is like two peas in a pod. Magnesium bromide (MgBr2) is indeed soluble in water, so it will form a precipitate.

b) 0.10 M NH4Br - Ah, now we're talking. Ammonium bromide (NH4Br) is also soluble in water, so it won't form a precipitate. It's a friendly solution.

c) 0.10 M FeCl2 - Uh-oh, this answer is a bit rusty. Iron(II) chloride (FeCl2) is not as soluble in water, so it will form a precipitate.

So, the answer is b) 0.10 M NH4Br, because it won't form a precipitate when added to 0.10 M NaOH. Just think of it as the life of the party, always mingling without causing any drama.

To determine which of the solutions will not form a precipitate when added to 0.10M NaOH, we need to check the solubility rules. Here are the step-by-step explanations for each solution:

a) 0.10 M MgBr2:
According to the solubility rules, bromides (Br-) are usually soluble, except when combined with silver (Ag+), lead (Pb2+), or mercury (Hg2+). Therefore, MgBr2 is soluble and will not form a precipitate when added to NaOH.

b) 0.10 M NH4Br:
Ammonium salts (NH4+) are usually soluble, so NH4Br will also be soluble and will not form a precipitate when added to NaOH.

c) 0.10 M FeCl2:
Chlorides (Cl-) are usually soluble, except when combined with silver (Ag+), lead (Pb2+), or mercury (Hg2+). Therefore, FeCl2 is soluble and will not form a precipitate when added to NaOH.

Therefore, all of the given solutions (a), b), and c)) will not form a precipitate when added to 0.10M NaOH.

To determine which of the given solutions will not form a precipitate when added to 0.10M NaOH, we need to identify the compounds that are insoluble in NaOH.

The solubility rules indicate that hydroxides of group 1 metals (alkali metals) are soluble, so we can eliminate NaOH itself as an insoluble compound.

Next, we need to examine the given compounds and determine if they will form insoluble hydroxides when mixed with NaOH:

a) 0.10 M MgBr2: To determine if MgBr2 will form an insoluble hydroxide, we need to check the solubility rules. According to the rules, most hydroxides of transition metals, such as Mg(OH)2, are only slightly soluble. Therefore, MgBr2 has the potential to form an insoluble precipitate.

b) 0.10 M NH4Br: Ammonium salts, such as NH4Br, are generally soluble in water and do not form insoluble hydroxides. Therefore, NH4Br will not form a precipitate when added to NaOH.

c) 0.10 M FeCl2: Similar to MgBr2, FeCl2 is a transition metal compound. Therefore, we need to check the solubility rules for hydroxides of transition metals. According to the rules, most hydroxides of transition metals are only slightly soluble. Therefore, FeCl2 has the potential to form an insoluble precipitate.

Based on these considerations, the solution that will not form a precipitate when added to 0.10M NaOH is b) 0.10 M NH4Br.

To answer this question you must know the solubilities of substances. Here is a simplified list. For example, look on this chart and you will see that Fe(OH)2 is insoluble is soluble.

http://en.wikipedia.org/wiki/Solubility_table#P