(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

What is the concentration of ____ ions in a ___ mol dm^{-3}solution of ___?

2. Relevant equations

mass/molar mass = number of moles

volume/molar volume = number of moles

Not sure what other equations...

3. The attempt at a solution

I'm not understanding how this works. I have looked at the answers and I'm not really understanding it...

For example, the first question says:

"What is the concentration of iodide ions in a 1 mol dm^{-3}solution of KI?"

I really didn't know where to start, so I found that 1 mol dm^{-3}KI is the same as 166g dm^{-3}and didn't really know where to go from there. The answer sheet says the answer is 1. The next question asks the same thing, except for hydroxide ions in 0.01 mol dm^{-3}of Ba(OH)_{2}, and the answer given is 0.02. It looks like they are multiplying the concentration by the number of ions. I'm not sure I understand why (I kind of keep getting glimpses of understanding which quickly fades).

Some help understanding why the answer is 1 would be appreciated (and why the answer for the second one is 0.02).

I know that KI dissociates into K^{+}and I^{-}.

EDIT: Never mind, I think I understand now, I think it's because I was typing it I figured it out. Is it because;

KI -> K^{+}+ I^{-}

So concentration of K^{+}is 1M and I^{-}is 1M.

And for the second, Ba(OH)_{2}-> Ba^{2+}+ 2(OH)^{-}

Since this is for 1M we have to divide all by 100, so ratio is 0.01:0.02 so 0.02 is the answer.

I think I understand now, correct me if wrong, thanks.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Chemistry - Concentration of Ions

**Physics Forums | Science Articles, Homework Help, Discussion**