How to write balanced chemical reaction?

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SUMMARY

The discussion focuses on calculating the acid dissociation constant (Ka) for hydrogen sulfate (HSO4-) from a solution of sodium hydrogen sulfate (NaHSO4) dissolved in water. The balanced chemical reactions are established as NaHSO4 dissociating into Na+ and HSO4-, followed by the dissociation of HSO4- into H3O+ and SO42-. The calculated Ka value is 1.14 x 10-2 M, derived from the concentrations of H3O+, SO42-, and HSO4- based on the stoichiometry of the reactions.

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  • Understanding of acid-base chemistry and dissociation reactions
  • Knowledge of stoichiometry and concentration calculations
  • Familiarity with the concept of pH and its relation to hydronium ion concentration
  • Basic knowledge of ionic compounds and their behavior in aqueous solutions
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  • Study the concept of acid dissociation constants (Ka) and their significance in chemical equilibria
  • Learn about the ionization of weak acids and their equilibrium expressions
  • Explore the relationship between pH and concentration of hydronium ions in solutions
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Chemistry students, educators, and professionals involved in analytical chemistry or chemical education, particularly those focusing on acid-base equilibria and reaction stoichiometry.

HJKL
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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.
 
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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
 
Why do we write that first equation? And why do we get what we get in the second equation?
 
HJKL said:
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.
 
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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:
HJKL said:
[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-)?
 
Think about the dissociation stoichiometry.
 
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