What Is the Final Concentration of Hydroxide Ions After Reaction?

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Discussion Overview

The discussion revolves around calculating the final concentration of hydroxide ions (OH-) after a reaction between barium hydroxide (Ba(OH)2) and phosphoric acid (H3PO4). Participants explore the stoichiometry of the reaction, the limiting reactant, and the dissociation of Ba(OH)2 in solution.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the moles of hydroxide produced based on the limiting reactant, H3PO4, suggesting a final concentration of 0.05 M, but notes a discrepancy with an expected answer of 0.128 M.
  • Another participant questions the calculation of initial hydroxide and emphasizes the need to consider the dissociation of Ba(OH)2 in the final concentration of OH-.
  • There is a discussion about whether the remaining Ba(OH)2 contributes to the final concentration of hydroxide ions, with some participants asserting that it does.
  • One participant expresses confusion about the role of Ba(OH)2 in the reaction and the necessity of its dissociation to determine the final concentration of OH-.
  • Another participant clarifies that the remaining Ba(OH)2 will dissociate and contribute to the hydroxide concentration, prompting further calculations based on the stoichiometry of the reaction.
  • There is a calculation presented that suggests starting with 0.4 moles of OH- and adjusting for the amount consumed in the reaction to find the remaining hydroxide ions.

Areas of Agreement / Disagreement

Participants express differing views on how to calculate the final concentration of hydroxide ions, particularly regarding the contributions from the dissociation of Ba(OH)2 and the impact of the limiting reactant. The discussion remains unresolved with multiple competing views on the calculations involved.

Contextual Notes

Participants note the importance of considering both the stoichiometry of the reaction and the dissociation of Ba(OH)2 to accurately determine the final concentration of hydroxide ions. There are unresolved aspects regarding the calculations and assumptions made about the initial and remaining quantities of reactants.

Lori

Homework Statement



What is the final concentration of hydroxide (OH) ions when 2.00 L of 0.100 M Ba(OH)2 is combined with 1.00 L of 5.00*10^-3 M H3PO4 and allowed to react to completion?

[/B]

Homework Equations



Chemical Equation that's balanced is:

3Ba(OH)2 + 2H3PO4 --> Ba3(PO4)2 +6H20[/B]

The Attempt at a Solution



To start off, i found that H3PO4 is the limiting reactant so used the mols of that to calculate hydroxide ions.

Therefore, 0.005 mols H3PO4 * (6 mols OH/2 mols H3PO4) = 0.15 mols OH
so .15 / 3 liters = 0.05 M (but the answer is 0.128 M!)

Can someone help me![/B]
 
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Lori said:
Therefore, 0.005 mols H3PO4 * (6 mols OH/2 mols H3PO4) = 0.15 mols OH

Can you explain what have you calculated here?
 
Now that I think about it. I calculated the mols of initial hydroxide .
 
No. Initial hydroxide is 2 L of 0.1 M (times two).
 
Borek said:
No. Initial hydroxide is 2 L of 0.1 M (times two).
How can I calculate final concentration of hydroxide if I don't see hydroxide in the product side then?

I'm not sure what OH to calculate since I can use the mol ratio and the mols of PO4 to calculate OH ions too.
 
What can you tell about hydroxide if the acid was the limiting reagent?
 
Hi Lori,

Firstly, 0.005 times 3 equals 0.015.

Secondly, think about Ba(OH)2. Does the remaining Ba(OH)2 play a role in (OH) final concentration?
 
Borek said:
What can you tell about hydroxide if the acid was the limiting reagent?
There would be 0.015 mols because 0.005 x (3).
 
DoItForYourself said:
Hi Lori,

Firstly, 0.005 times 3 equals 0.015.

Secondly, think about Ba(OH)2. Does the remaining Ba(OH)2 play a role in (OH) final concentration?
Yes it does. It would become water right
 
  • #10
The remaining Ba(OH)2 will dissociate as follows (in water) : Ba(OH)2 ## \to ## Ba2+ + 2 OH-.

Hint: Find the remaining quantity of Ba(OH)2 and find the mols of OH from the reaction stoichiometry.
 
  • #11
DoItForYourself said:
The remaining Ba(OH)2 will dissociate as follows (in water) : Ba(OH)2 ## \to ## Ba2+ + 2 OH-.

Hint: Find the remaining quantity of Ba(OH)2 and find the mols of OH from the reaction stoichiometry.
I'm confused cause I thought that the chemical equation I gave is correct. Why is there suddenly dissociation for barium hydroxide??
 
  • #12
The chemical equation you gave is correct. But as you said, this equation does not give OH-.

In addition, the acid is the limiting reagent (see post #6), and thus some quantity of Ba(OH)2 will remain in the final solution (this quantity of Ba(OH)2 did not react with H3PO4). This remaining quantity will dissociate in the solution. The dissociation reaction is what you need to calculate the final concentration of OH-.
 
  • #13
DoItForYourself said:
The chemical equation you gave is correct. But as you said, this equation does not give OH-.

In addition, the acid is the limiting reagent (see post #6), and thus some quantity of Ba(OH)2 will remain in the final solution (this quantity of Ba(OH)2 did not react with H3PO4). This remaining quantity will dissociate in the solution. The dissociation reaction is what you need to calculate the final concentration of OH-.
So the only OH is coming from the dissociation?
 
  • #14
Yes, but you have to calculate the remaining quantity of Ba(OH)2 that dissociates, so you have to take into consideration both chemical equations.
 
  • #15
DoItForYourself said:
The chemical equation you gave is correct. But as you said, this equation does not give OH-.

In addition, the acid is the limiting reagent (see post #6), and thus some quantity of Ba(OH)2 will remain in the final solution (this quantity of Ba(OH)2 did not react with H3PO4). This remaining quantity will dissociate in the solution. The dissociation reaction is what you need to calculate the final concentration of OH-.
Oh I seee. So we started off with 0.4 mols of OH but the acid limits the OH so there will be OH ions left . 0.4 - 0.015 = .128. Ions left that dissociate and so divide by 3 is .128 M
 
  • #16
To be more precise, 0.015 out of the initial 0.4 moles were consumed (in the reaction you gave initially), so the remaining OH- are 0.385 moles. If you divide them by the volume, the result gives you the final concentration, right.
 

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