determining cations or anions

[Emily_Stenzler_2H]

are we supposed to know how to do problems like J17a (in the textbook) on the final? I am very confused about how this would be solved :/

In each of the following salts, either the cation or the anion is a weak acid or a weak base in water. Write the chemical equation for the proton transfer reaction of this cation or anion with water: (a) NaC6H5O

[Taha 2D]

You can tell it is basic because the cation is Na and that is a specatator ion. the proton transfer is just the anion pulling a hydrogen from water and leaving OH-

[JoshMoore2B]

are we supposed to know how to do problems like J17a (in the textbook) on the final? I am very confused about how this would be solved :/

In each of the following salts, either the cation or the anion is a weak acid or a weak base in water. Write the chemical equation for the proton transfer reaction of this cation or anion with water: (a) NaC6H5O

Generally, the best way to look at these types of questions is to see what the cation and anion are and what acid and base they might create.

In this case, the cation is Na+ and the anion is C6H5O-. As we know, Na+ is involved with the strong base NaOH. What this means is that when the Na+ ion is in the solution, the compound it will want to make is NaOH. However, because NaOH is a strong base, its presence in solution is unfavorable (i.e. it wants to fully disassociate into Na+ and OH-). This means that the Na+ won't affect the pH, because it isn't favorable for NaOH to exist in solution.

However, for C6H5O-, this compound is not associated with a strong acid. This means that it is somewhat favored in solution for an acid (presumably C6H6O) to exist in its associated state, and it is somewhat favored for that acid to exist in the H+ and C6H5O- disassociated state. So, when NaC6H5O disassociates in water and releases the two ions, the Na+ won't be able to form NaOH, but the C6H5O- will form some C6H6O. This means that the anion has taken some H+ away from some water molecules in solution, meaning that H2O has had an H taken from it, leaving behind OH-. This means C6H5O- has increased the pH of the solution, because it has increased the concentration of OH-.

This process is applicable to all other salts. I hope this helped!

P.S. Some salts are neutral in water, like NaCl, because Na+ is associated with NaOH and Cl- is associated with HCl, both of which are strong, meaning the ions won't affect anything, leaving the solution neutral.

[SelenaDahabreh1D]

are we supposed to know how to do problems like J17a (in the textbook) on the final? I am very confused about how this would be solved :/

In each of the following salts, either the cation or the anion is a weak acid or a weak base in water. Write the chemical equation for the proton transfer reaction of this cation or anion with water: (a) NaC6H5O

Generally, the best way to look at these types of questions is to see what the cation and anion are and what acid and base they might create.

In this case, the cation is Na+ and the anion is C6H5O-. As we know, Na+ is involved with the strong base NaOH. What this means is that when the Na+ ion is in the solution, the compound it will want to make is NaOH. However, because NaOH is a strong base, its presence in solution is unfavorable (i.e. it wants to fully disassociate into Na+ and OH-). This means that the Na+ won't affect the pH, because it isn't favorable for NaOH to exist in solution.

However, for C6H5O-, this compound is not associated with a strong acid. This means that it is somewhat favored in solution for an acid (presumably C6H6O) to exist in its associated state, and it is somewhat favored for that acid to exist in the H+ and C6H5O- disassociated state. So, when NaC6H5O disassociates in water and releases the two ions, the Na+ won't be able to form NaOH, but the C6H5O- will form some C6H6O. This means that the anion has taken some H+ away from some water molecules in solution, meaning that H2O has had an H taken from it, leaving behind OH-. This means C6H5O- has increased the pH of the solution, because it has increased the concentration of OH-.

This process is applicable to all other salts. I hope this helped!

P.S. Some salts are neutral in water, like NaCl, because Na+ is associated with NaOH and Cl- is associated with HCl, both of which are strong, meaning the ions won't affect anything, leaving the solution neutral.

So if the anion/cation is associated with a strong acid/base, then it is not favorable in our solution and we will leave it out of the reaction?

[Tanya Nguyen 1B]

You can tell it is basic because the cation is Na and that is a specatator ion. the proton transfer is just the anion pulling a hydrogen from water and leaving OH-

What is a spectator ion and how are we supposed to identify it? Was it ever mentioned in one of the lectures?

[Taha 2D]

You can tell it is basic because the cation is Na and that is a specatator ion. the proton transfer is just the anion pulling a hydrogen from water and leaving OH-

What is a spectator ion and how are we supposed to identify it? Was it ever mentioned in one of the lectures?

it is mentioned somewhere in the lectures, maybe in 14a not sure. but it is basically an ion which doesnt take part in the reaction. so itll be there in the same quantities as before and after the reaction ( Na fully disassociates so its a spectator ion)