Molarity Calculation for Ba(NO3)2 Solution

  • Thread starter Shackleford
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In summary: I haven't done a limiting reagent problem in well over two years. EOT?End Of Thread, I believe. Do you understand the answer?In summary, the homework statement is trying to find the molarity of the solution (salt and any excess reagent after reaction of 250 mL of 0.05 M Ba(OH)2 with 200 mL of 0.08 M HNO3). The Attempt at a Solution found that only .008 moles of barium nitrate will be formed with .017 M of nitric acid as the limiting reagent.
  • #1
Shackleford
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Homework Statement



Find the molarity of the solution (salt and any excess reagent after reaction of 250 mL of 0.05 M Ba(OH)2 with 200 mL of 0.08 M HNO3.

Homework Equations


The Attempt at a Solution



I am drawing a blank. Here's the balanced equation

Ba(OH)2 + 2HNO3 ===> 2H2O + Ba(NO3)s

Ba(OH) .0125 moles

2HNO3 .016 moles
 
Last edited:
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  • #2
Simple stoichiometry, this is a limiting reagent question. Your reaction equation is almost OK (chack barium nitrate formula, but I suppose that's a typo) and is a correct first step to solution.
 
  • #3
Why can I not edit my own damn post?
 
  • #4
You can for 30 minutes, or something like that.
 
  • #5
Borek said:
You can for 30 minutes, or something like that.

Ok, well, Ba(NO3)2, correct?
 
  • #6
Exactly.

Ba(OH)2 + 2HNO3 -> Ba(NO3)2 + 2H2O

Now try the limiting reagent approach.
 
  • #7
Borek said:
Exactly.

Ba(OH)2 + 2HNO3 -> Ba(NO3)2 + 2H2O

Now try the limiting reagent approach.

I used the number of moles not grams to find the limiting reagent, which is nitric acid. Only .008 moles of barium nitrate will be formed.

.008 moles/.450 L = .017 M
 
  • #8
0.008 Ba(NO3)2 is a correct intermediate result. But that's not what you were asked about.
 
  • #9
Borek said:
0.008 Ba(NO3)2 is a correct intermediate result. But that's not what you were asked about.

Borek
--
equation balancer and stoichiometry calculator

I know. Check my previous post. I put the molarity of the solution.
 
  • #10
It is not all. Excess reagent is still present in the solution in unchanged form.
 
  • #11
Borek said:
It is not all. Excess reagent is still present in the solution in unchanged form.

Ah. I forgot about that.
 
Last edited:
  • #12
You should list every ion and its concentration separately. EOT for me.
 
  • #13
  • #14
End Of Thread, I believe. Do you understand the answer?
 

Related to Molarity Calculation for Ba(NO3)2 Solution

What is the molarity of Ba(NO3)2 0.14M?

The molarity of a solution is defined as the number of moles of solute per liter of solution. In this case, the molarity of Ba(NO3)2 0.14M means that there are 0.14 moles of Ba(NO3)2 dissolved in every liter of solution.

How do you calculate the molarity of a solution?

To calculate the molarity of a solution, you divide the number of moles of solute by the volume of the solution in liters. This can be represented by the formula M = moles of solute / liters of solution.

Why is molarity an important measurement in chemistry?

Molarity is an important measurement in chemistry because it allows us to accurately represent the concentration of a solution. This is useful in many applications, such as determining the correct amount of a substance to use in a reaction or understanding how different concentrations of a solution can affect a system.

How can you prepare a 0.14M solution of Ba(NO3)2?

To prepare a 0.14M solution of Ba(NO3)2, you would need to dissolve 0.14 moles of Ba(NO3)2 in enough water to make a final volume of 1 liter. This can be done by measuring out the appropriate amount of solid Ba(NO3)2 and adding it to a volumetric flask, then adding water to reach the 1 liter mark.

What factors can affect the molarity of a solution?

The molarity of a solution can be affected by factors such as temperature, pressure, and the presence of other solutes. Changes in these factors can alter the volume or number of moles of solute in a solution, thus changing the molarity. It is important to control for these factors when preparing and measuring molarity in experiments.

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