How Many Moles of BaCO3 Dissolve in Water Given Kc?

In summary: I was dividing by 5L thinking I needed to isolate moles but I actually needed to keep it in the denominator and then multiply the whole equation by 5L to get moles by itself.
  • #1
Not a Wrench
22
1

Homework Statement


Excess BaCO3 is placed in a container containing 5L of pure water. Given the following info:
BaCO3 (s) ↔ Ba2+ + CO3 2- and Kc = 5.7E-6
how many moles of BaCO3 dissolve?

Homework Equations


Kc=(Ba2+ )(CO3 2- ) = 5.7E-6

The Attempt at a Solution


solved for x to get .002387 moles/5L which is incorrect. Dividing by 5L gives me .000477 moles. The correct answer is .012 and I am unsure what I have done wrong. It seemed simple enough. Could someone tell me what to do?
 
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  • #2
0.002387 is not number of moles.
 
  • #3
Borek said:
0.002387 is not number of moles.
I forgot to add my last step but I know. It's the moles per liter. If I divide by 5 it gives me .000477 moles which is also not correct.
 
  • #4
Not a Wrench said:
.000477 moles

No, it is not moles. Check your units.

Randomly juggling numbers is NEVER a good approach to solving problems.
 
  • #5
Borek said:
No, it is not moles. Check your units.

Randomly juggling numbers is NEVER a good approach to solving problems.
Then what is x supposed to be? I was under the impression that the number by which it BaCO3 decreases is equal to the x that each of the products increases by.
 
  • #6
So you are solving for x not knowing what it is? That's randomly juggling numbers, I told you it doesn't work.

You have calculated something in your first post. What it was? Can you post your whole solution?
 
  • #7
Borek said:
So you are solving for x not knowing what it is? That's randomly juggling numbers, I told you it doesn't work.

You have calculated something in your first post. What it was? Can you post your whole solution?
Borek said:
So you are solving for x not knowing what it is? That's randomly juggling numbers, I told you it doesn't work.

You have calculated something in your first post. What it was? Can you post your whole solution?
Yeah, all I did was calculate 5.7E-6 = x2 and solved for x which got me to .002387. Isn't its units M/5L because the container contains 5L of water?
 
  • #8
Not a Wrench said:
Isn't its units M/5L

No, that's not how units work. Units don't contain numbers (or, if anything, we assume all numbers to be equal to 1). Besides, typically by M we denote molar concentration (which has its own units, mol/L ← an important hint here!), not number of moles.

Not a Wrench said:
5.7E-6 = x2

What is x intended to mean in this equation?
 
  • #9
Borek said:
No, that's not how units work. Units don't contain numbers (or, if anything, we assume all numbers to be equal to 1). Besides, typically by M we denote molar concentration (which has its own units, mol/L ← an important hint here!), not number of moles.
What is x intended to mean in this equation?
The Kc equation contains (Ba2+ )(CO32- ) and they both gain x per the ice chart so therefore Kc = x2
 
  • #10
And what you mean by (Ba2+) (or (CO32-))?
 
  • #11
Borek said:
And what you mean by (Ba2+) (or (CO32-))?
Those are the products in the equilibrium. But I figured it out, anyways. I was dividing by 5L of water when I should have been multiplying by 5L because I needed to eliminate that 5L in the denominator and get moles by itself!
 
  • #12
Not a Wrench said:
Those are the products in the equilibrium.

These are CONCENTRATIONS. And concentration has units of mol/L. Per ONE liter, not per 5 liters.

When you divided by 5 L you got the result in mol/L2 units - these don't make sense, but you have never checked so you were not able to spot the mistake.
 
  • #13
Borek said:
These are CONCENTRATIONS. And concentration has units of mol/L. Per ONE liter, not per 5 liters.

When you divided by 5 L you got the result in mol/L2 units - these don't make sense, but you have never checked so you were not able to spot the mistake.
I know, thank you for the hint.
 

1. What is chemical equilibrium?

Chemical equilibrium is a state in which the forward and reverse reactions of a chemical reaction occur at equal rates, resulting in no net change in the concentration of reactants and products.

2. How is equilibrium achieved in a chemical reaction?

Equilibrium is achieved when the concentrations of reactants and products reach a balance, meaning that the rates of the forward and reverse reactions are equal. This can occur naturally over time or can be reached through changes in temperature, pressure, or concentration.

3. What factors can affect chemical equilibrium?

Temperature, pressure, and concentration are the main factors that can affect chemical equilibrium. Changes in these factors can shift the equilibrium in favor of either the reactants or products.

4. How is equilibrium represented in a chemical equation?

Equilibrium is represented by a double arrow (⇌) between the reactants and products in a chemical equation. This indicates that the reaction can proceed in either direction.

5. Can equilibrium be disturbed?

Yes, equilibrium can be disturbed by changes in temperature, pressure, and concentration. These changes can shift the equilibrium in favor of either the reactants or products, resulting in a new equilibrium state.

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