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Topic: Ions Present  (Read 8676 times)

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Offline Violet89

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Ions Present
« on: May 01, 2013, 09:44:41 PM »
The extent of fragmentation for ionic compounds can be derived by comparing the molality expected if no dissociation with what is actually obtained from experiment. When 0.278 g of K3Fe(CN)6 (molar mass 329.25 g/mol) is dissolved in 100.0 g of water, the freezing point is found to be -0.070 °C. How many ions are present for each formula unit of K3Fe(CN)6 dissolved? (Kf for water is 1.86 K/m).

The answer is 5.

In water, the complex salt dissociates: K4Fe(CN)6 ==> 4K^+ + Fe(CN)6^-4

When K4Fe(CN)6 dissociates in water, there are 5 ions. Is this all I need to know for this problem, or these types of problems when asking for ions present? I'm confused. I'd appreciate any help. :-)

Offline Hunter2

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Re: Ions Present
« Reply #1 on: May 02, 2013, 01:12:14 AM »
What you talking about? In beginning its K3[Fe(CN)6] and below its  K4[Fe(CN)6].

Offline Violet89

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Re: Ions Present
« Reply #2 on: May 02, 2013, 01:28:47 AM »
What you talking about? In beginning its K3[Fe(CN)6] and below its  K4[Fe(CN)6].

Oops, yes, it's supposed to be 4 in the problem. That's what I'm talking about...

Offline Borek

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Re: Ions Present
« Reply #3 on: May 02, 2013, 03:11:30 AM »
You should use the information given to calculate Van 't Hoff factor.
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Offline Violet89

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Re: Ions Present
« Reply #4 on: May 02, 2013, 05:35:37 AM »
You should use the information given to calculate Van 't Hoff factor.

Thanks.

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