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Topic: Solving for concentration - ionic equilibrium  (Read 1181 times)

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Offline ch3mical

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Solving for concentration - ionic equilibrium
« on: August 03, 2021, 12:04:35 PM »
Hi,

I have attempted an answer to the question below. Please let me know if my reasoning is correct. Thanks.

Q) You have a 1L solution of barium chromate, where the concentration of chromate ions is 110nM. How many mols of barium chloride should be added in order to reduce the concentration of the chromate ions to 12nM?

My attempt
Equation for barium chromate in solution:
BaCrO4 ::equil:: Ba2+ + CrO42-

When we add barium chloride, the balanced equation will be:
Ba2+ + CrO42- + BaCl2 ::equil:: BaCrO4 + BaCl2-

We are told that [CrO42-] = 110nM, and we want to reduce it to 12nM. We have 1L, so that means we want to go from 110 mols down to 12 mols (mols = M×volume). This is a difference of 98 mols. From the equations above, 1 mol of barium chloride eliminates 1 mol of chromate ions. Therefore the mols of barium chloride required to consume 98 mols of chromate ions are 98 mols of barium chloride. So, my answer is 98 mols of BaCl2.

Are my steps of reasoning correct? I don't know what the correct answer is.

Offline Orcio_Dojek

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Re: Solving for concentration - ionic equilibrium
« Reply #1 on: August 03, 2021, 04:08:55 PM »
I doubt that 98 moles (20 kg) of BaCl2 can be dissolved in less than 1 liter of water.

But - does this problem does not need use of the solubility product ?

Offline Borek

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Re: Solving for concentration - ionic equilibrium
« Reply #2 on: August 03, 2021, 04:14:06 PM »
We are told that [CrO42-] = 110nM, and we want to reduce it to 12nM. We have 1L, so that means we want to go from 110 mols down to 12 mols (mols = M×volume).

Nope.

n in nM stands for nano, metric prefix used with units.

(and - as Orcio already suggested - this question is about solubility product)
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Offline ch3mical

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Re: Solving for concentration - ionic equilibrium
« Reply #3 on: August 03, 2021, 04:26:47 PM »
Thank you both. Ah yes, I meant 98 nano mols.

Could you please show me how to apply the solubility product equation in this case?

Offline ch3mical

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Re: Solving for concentration - ionic equilibrium
« Reply #4 on: August 03, 2021, 05:20:38 PM »
Here's my thinking about how I would apply the solubility product.

Ksp of barium chromate = [Ba2+] [CrO42-]

We are given that [CrO42-] = 11μM, therefore [Ba2+] is also 11μM, and Ksp=121×10-12.

The question states that we add barium chloride in order to get a new [CrO42-] of 12nM. Given that we've calculated Ksp=121×10-12, we can substitute in the new [CrO42-], and solve for [Ba2+] to get 1×10-20M. I have assumed that one mole of barium chloride eliminates one mole of the barium ions.

Thoughts?

Offline Orcio_Dojek

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Re: Solving for concentration - ionic equilibrium
« Reply #5 on: August 03, 2021, 05:31:00 PM »
Quote
The question states that we add barium chloride in order to get a new [CrO42-] of 12nM. Given that we've calculated Ksp=121×10-12, we can substitute in the new [CrO42-], and solve for [Ba2+] to get 1×10-20M. I have assumed that one mole of barium chloride eliminates one mole of the barium ions.
It does not "eliminate" anything. Ksp has the value 1,21 x 10-10 and concentration of the CrO42- should be smaller than 12 x 10-9 M, so concentration of the Ba2+ should be bigger than.... ?

Offline ch3mical

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Re: Solving for concentration - ionic equilibrium
« Reply #6 on: August 03, 2021, 05:45:53 PM »
It does not "eliminate" anything. Ksp has the value 1,21 x 10-10 and concentration of the CrO42- should be smaller than 12 x 10-9 M, so concentration of the Ba2+ should be bigger than.... ?
I think I see what you are saying. Ksp is a constant, so if the chromate ion concentration goes down, then the barium ion concentration must go up to Ksp/[CrO42-] = 10 mM. So, the required concentration of barium ions is this value. In my post below, I try to work back from this to the required moles of barium chloride...
« Last Edit: August 03, 2021, 05:59:20 PM by ch3mical »

Offline ch3mical

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Re: Solving for concentration - ionic equilibrium
« Reply #7 on: August 03, 2021, 05:57:49 PM »
Is the following approach correct?

We started off with 11uM of barium ions, which equated to 11μmol in our 1L solution.

Based on the working out above, we end up with 10mM of barium ions after adding the barium chloride. This means we now have 10mmols. The difference in mols after adding barium chloride is 10mmols - 11μmols ≈ 9.99mmols. This represents the amount of barium chloride we needed to add, because 1 mol of barium chloride reacts with 1 mol of chromate ions. I am not sure if my answer is correct.

Offline Orcio_Dojek

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Re: Solving for concentration - ionic equilibrium
« Reply #8 on: August 04, 2021, 01:33:32 AM »
110 nM is 0,11 not 11 uM, but the rest is aprox. true (concentration of Ba2+ should be bigger than 10 mM) - of course if dissolving 2,08 g BaCl2 will not change volume of 1 liter solution.

Offline ch3mical

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Re: Solving for concentration - ionic equilibrium
« Reply #9 on: August 04, 2021, 04:52:45 PM »
110 nM is 0,11 not 11 uM, but the rest is aprox. true (concentration of Ba2+ should be bigger than 10 mM) - of course if dissolving 2,08 g BaCl2 will not change volume of 1 liter solution.
Thank you

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