# 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?

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## 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]

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).

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?

What can you tell about hydroxide if the acid was the limiting reagent?
There would be 0.015 mols because 0.005 x (3).

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

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.

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??

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-.

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?

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.

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

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.