Topic 6D Exercise 11

[Vivian Leung 1C]

Hi everyone,

I am a little confused as to how the equilibrium equation for 11e and 11f is created. Could someone walk me through how to get there?

[Vivian Leung 1C]

There is likely an error in the solutions/textbook seeing as the molecules are not the same between the two.

I see. Thank you. I just realized that the picture I sent of the solution was cut off a little bit so the answer is what I attached here instead. Would this be the correct answer? And if so, how do I get there?

[Chem_Mod]

As a correction and further explanation to my previous response:

For AlCl3, the compound dissociates into Al3+ and Cl-. Since Cl- is the conjugate base of a strong acid, Cl- will not be very reactive with water because it is an extremely weak base. Therefore, we are left with only Al3+ to deal with.

Al3+ is a metal cation, which can serve as the central metal ion for a coordination compound with H2O. Al can bond 6 times with H2O.

Al(H2O)6 3+ can lose a hydrogen from one of the H2Os and give it to water, leading to the answer key's products.

The same concepts apply for part (f) as NO3- is also the conjugate base of a strong acid.

[Vivian Leung 1C]

As a correction and further explanation to my previous response:

For AlCl3, the compound dissociates into Al3+ and Cl-. Since Cl- is the conjugate base of a strong acid, Cl- will not be very reactive with water because it is an extremely weak base. Therefore, we are left with only Al3+ to deal with.

Al3+ is a metal cation, which can serve as the central metal ion for a coordination compound with H2O. Al can bond 6 times with H2O.

Al(H2O)6 3+ can lose a hydrogen from one of the H2Os and give it to water, leading to the answer key's products.

The same concepts apply for part (f) as NO3- is also the conjugate base of a strong acid.

I see, thank you!