How to write balanced chemical reaction?

  • Thread starter HJKL
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  • #1
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Main Question or Discussion Point

Hi!
I've got this problem:
1,20 g of NaHSO4 is dissolved in water to 1,0 liters and pH=2,194. What is the acid dissociation constant for HSO-4?

I assume I have to start with writing the balanced reaction, but I don't get how I do that.
I know that I'm supposted to start with NaHSO4 + H2O , but not the rest.
 

Answers and Replies

  • #2
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Solution:
NaHSO4 ---> Na+ + HSO4-
HSO4- + H2O <--> H3O+ + SO42-

Moles of NaHSO4 : 1,2g / 120 g/mol = 0.01 mol

[H3O+] = 10-2.194
[SO42- = 10-2.194
[HSO4- = 0.01 - 10-2.194

Acid dissociation constant:

ka = 10-2.194 * 10-2.194 / 0.01-102.194
ka = 1.14 * 10-2 M
 
  • #3
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Why do we write that first equation? And why do we get what we get in the second equation?
 
  • #4
Merlin3189
Homework Helper
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Why do we write that first equation? And why do we get what we get in the second equation?
I know that I'm supposted to start with NaHSO4 + H2O ,
When the sodium hydrogen sulphate dissolves in water, it ionises and that is your first equation.
The hydrogen sulphate ion further dissociates in water giving your second equation.

The ionisation of the salt is complete in dilute solution, but the dissociation is not. It is the second part for which you are calculating the dissociation constant, so that is the equation you need.

You get what you get in that equation because that is what happens!
Sodium hydrogen sulphate is an ionic compound of Na+ and HSO4- ions, so dissociates completely in dilute aqueous solution.
The hydrogen sulphate ion is itself covalently bonded, but is enticed to lose a proton to water and dissociate to some degree. On the other hand, the sulphate ions can also accept a proton to reform hydrogen sulphate ions. You are calculating the balance between these two contrary reactions.
The sulphate ion is very stable and doesn't breakdown further in water.
 
  • #5
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Ok, thanks. So when I've got my equation and I then I'm supposted to find the constant. Why do I do this:
[H3O+] = 10-2.194
[SO42- = 10-2.194
[HSO4- = 0.01 - 10-2.194
I find the concentration of H3O, SO42- and HSO4-. But why is the concentration of HSO4- equal to 0.01 (moles of NaHSO4) - 10-2.194 (concentration of H3O and SO42-)?
 
  • #6
Borek
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Think about the dissociation stoichiometry.
 

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