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Chemistry 13 Online
OpenStudy (anonymous):

NiCl3 is a strong electrolyte. Determine the concentration of each of the individual ions in a 0.600 M NiCl3 solution.

OpenStudy (rushwr):

What's ur answer ! I'l just check it !!!

OpenStudy (anonymous):

Well my question is on how to get it. How do the charges of the individual ions affect the concentration?

OpenStudy (anonymous):

I don't know how to find the concentration of the ions themselves

OpenStudy (rushwr):

is that all the data given ?/

OpenStudy (anonymous):

Yes and that Na is a +3 charge and Cl is -1

OpenStudy (rushwr):

Okai I'm not soo sure with this though but I'll try. First we will write the dissociation equation for this \[NiCl _{3} \rightarrow Ni ^{3+} + 3Cl ^{-}\]

OpenStudy (anonymous):

okay

OpenStudy (rushwr):

So I'm just sayn try using the values I'm giving. Ni^3+ = 0.6M Cl^- = 1.8M

OpenStudy (rushwr):

Did u check ?

OpenStudy (anonymous):

Yes! It's right! How did you get those values though?

OpenStudy (rushwr):

look at the dissociation equation. The moles of NiCl3 is equal to the moles of Ni^3+ right? If we consider the same volume the concentration becomes equal. And we can see the moles of Cl^-1 is thrice as of Ni^3+ so u multiply 0.6 by 3 = 1.8M

OpenStudy (rushwr):

I'm sorry I didn't see that comment coming !

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