# What Is the Final Concentration of Hydroxide Ions After Reaction?

• Lori
So the concentration is 0.385 / 3 = 0.128 mol/L.In summary, the final concentration of hydroxide (OH) ions can be calculated by taking into consideration the limiting reagent, which in this case is H3PO4. The remaining Ba(OH)2 will dissociate in the solution and contribute to the final concentration of OH-. Therefore, the final concentration is 0.128 mol/L.
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]

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

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.

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??

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-.

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?

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.

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

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.

## 1. What is an acid/base stoichiometry problem?

An acid/base stoichiometry problem is a type of chemistry problem that involves determining the amount of reactants and products involved in an acid-base reaction. It requires the use of stoichiometric calculations to determine the quantities of substances involved in the reaction.

## 2. What is the purpose of solving an acid/base stoichiometry problem?

The purpose of solving an acid/base stoichiometry problem is to accurately determine the amount of reactants that are needed to completely react with each other, as well as the amount of products that will be formed. This information is crucial in order to perform experiments and predict the outcome of reactions.

## 3. How do you approach an acid/base stoichiometry problem?

The first step in approaching an acid/base stoichiometry problem is to write out the balanced chemical equation for the reaction. Then, use the given information to determine the moles of the known substance. Next, use the mole ratio from the balanced equation to calculate the moles of the unknown substance. Finally, convert the moles to the desired units.

## 4. What are some common mistakes when solving an acid/base stoichiometry problem?

Some common mistakes when solving an acid/base stoichiometry problem include forgetting to balance the chemical equation, using incorrect units, and not properly converting between units. It is important to carefully follow the steps and double check the calculations to avoid these mistakes.

## 5. How can I check my answer for an acid/base stoichiometry problem?

To check your answer for an acid/base stoichiometry problem, you can use the law of conservation of mass. This law states that matter cannot be created or destroyed, so the total mass of the reactants should equal the total mass of the products. Additionally, you can also double check your calculations and units to ensure they are correct.

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