Need Help with Acid/Base Stoic problem

[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]

 

[Borek]

Therefore, 0.005 mols H3PO4 * (6 mols OH/2 mols H3PO4) = 0.15 mols OH

Can you explain what have you calculated here?

 

[Lori]

Now that I think about it. I calculated the mols of initial hydroxide .

 

[Borek]

No. Initial hydroxide is 2 L of 0.1 M (times two).

 

[Lori]

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.

 

[Borek]

What can you tell about hydroxide if the acid was the limiting reagent?

 

[DoItForYourself]

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?

 

[Lori]

What can you tell about hydroxide if the acid was the limiting reagent?

There would be 0.015 mols because 0.005 x (3).

 

[Lori]

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

 

[DoItForYourself]

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.

 

[Lori]

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??

 

[DoItForYourself]

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^-.

 

[Lori]

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?

 

[DoItForYourself]

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.

 

[Lori]

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

 

[DoItForYourself]

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.