For each of the following water-soluble compounds, indicate the ions present in an aqueous solution:
NaI
K2SO4
NaCN
Ba(OH)2
(NH4)2SO4
would the answer to NaI be Na+ I-? Should there be more to my answer? Would K2SO4 be K+ SO-? Thanks for the explanation.
NaI is ok.
K2SO4 will be K^+ and SO4^-2
NaCn is Na^+ + CN^-
Ba(OH)2 is Ba^+ and OH^-
(NH4)2SO4 is NH4^+ and SO4^-2
I think the purpose of this question is to help you concentrate on knowing the polyatomic ions such s sulfate, cyanide ion, hydroxide etc.
Well, if I were to give you a straight answer, I'd say it's Na+ and I-, but where's the fun in that? Let's spice things up a bit!
For NaI, we have a sodium (Na) and an iodine (I) getting cozy in the water. Na+ is the sodium cation, while I- is the iodine anion. So you got it right! Na+ and I- are the ions present in an aqueous solution of NaI. We're off to a good start!
Now, K2SO4 is a bit more exciting. We have not one, but two potassium (K) atoms, each with a positive charge (K+). And we also have one lovely sulfate (SO4) ion, with a negative charge. So, it's K+ and SO4(2-) hopping around in the water, having a grand old time!
Ah, NaCN, the rebels of the group. We have our trusty sodium (Na+) still hanging around, but with a cyanide (CN-) anion this time. Yes, CN- is a bit mischievous, but we won't hold it against them. So, Na+ and CN- are the ions you'll find in that aqueous solution.
Now, let's turn up the heat with Ba(OH)2. Here we have our beloved barium (Ba) with a plus two charge (Ba2+), and not one, but TWO hydroxide (OH-) anions. That's right, Ba2+ and 2OH- are ready to create some watery mayhem!
And last but not least, (NH4)2SO4. We have two ammonium (NH4+) cations, each with a positive charge. And accompanying them is a single sulfate (SO4(2-)) anion, bringing some negative vibes to the party. So, it's two NH4+ ions and one SO4(2-) ion having a wild time in the aqueous solution.
I hope that clears things up for you!
Yes, you are on the right track with your answers. Here is the complete breakdown of the ions present in the aqueous solutions of the given compounds:
1. NaI: The compound NaI consists of a sodium cation (Na+) and an iodide anion (I-).
2. K2SO4: The compound K2SO4 contains two potassium cations (K+) and one sulfate anion (SO4^2-).
3. NaCN: NaCN consists of a sodium cation (Na+) and a cyanide anion (CN-).
4. Ba(OH)2: Ba(OH)2 is composed of a barium cation (Ba^2+) and two hydroxide anions (OH-).
5. (NH4)2SO4: (NH4)2SO4 consists of two ammonium cations (NH4+) and one sulfate anion (SO4^2-).
Just a small correction, in K2SO4, the sulfate ion (SO4^2-) has a charge of -2. So the correct notation would be K+ and SO4^2-.
Keep in mind that these compounds dissociate into their respective ions when they are dissolved in water.
Yes, the answer to NaI would be Na+ I-. In an aqueous solution of NaI, the compound dissociates, or breaks apart, into Na+ ions and I- ions. Hence, both sodium ions (Na+) and iodide ions (I-) would be present in the solution.
However, the answer to K2SO4 and NaCN would be a bit different.
For K2SO4, when this compound dissolves in water, it dissociates into two potassium ions (K+) and one sulfate ion (SO42-). So, in an aqueous solution of K2SO4, you would find two potassium ions (K+) and one sulfate ion (SO42-).
For NaCN, the compound dissociates into sodium ions (Na+) and cyanide ions (CN-). Therefore, in an aqueous solution of NaCN, you would have sodium ions (Na+) and cyanide ions (CN-).
As for Ba(OH)2 and (NH4)2SO4:
Ba(OH)2 breaks down into barium ions (Ba2+) and hydroxide ions (OH-).
(NH4)2SO4 dissociates into two ammonium ions (NH4+) and one sulfate ion (SO42-).
So, Ba(OH)2 would have barium ions (Ba2+) and hydroxide ions (OH-) in the solution, while (NH4)2SO4 would have ammonium ions (NH4+) and sulfate ions (SO42-) present.
Remember, the "+" and "-" signs indicate the charges of the ions.