Please help Calculate the % hydrolysis for 1 M Na2CO3 if [OH-] = 5.4x10^-(4).

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SUMMARY

The discussion focuses on calculating the percentage hydrolysis of a 1 M sodium carbonate (Na2CO3) solution given an hydroxide ion concentration of [OH-] = 5.4x10^(-4). The relevant hydrolysis reactions include the dissociation of carbonate ions (CO3^(2-)) into bicarbonate ions (HCO3^(-1)) and hydroxide ions. The equilibrium constants for the reactions involving carbonic acid (H2CO3) are provided, specifically Ka1 = 4.3x10^(-7) and Ka2 = 5.6x10^(-11). The calculation of Kb values for bicarbonate and carbonic acid is also discussed, leading to the determination of the theoretical percentage hydrolysis.

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  • Understanding of hydrolysis reactions in aqueous solutions
  • Familiarity with equilibrium constants (Ka and Kb)
  • Knowledge of the dissociation of carbonic acid (H2CO3)
  • Ability to perform calculations involving concentrations and equilibrium
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  • Learn how to calculate Kb from Ka values using the formula Kb = Kw/Ka
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Calculate the % hydrolysis for 1 M Na2CO3 if [OH-] = 5.4x10^-(4)
 
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Done.

Ignoring template and not showing any effort will lead you nowhere.
 
Borek said:
Done.

Ignoring template and not showing any effort will lead you nowhere.

Huh? But I asked my question. This is the only information I was given.

No effort? But I'm stuck! I'm showing some effort by asking for some help here on ONE problem!
 

Homework Statement



Calculate the % hydrolysis for 1 M Na2CO3 if [OH-] = 5.4x10^-(4).

Not sure what to put for the other fields.
 
Do you know any reaction equations? What is hydrolysis?
 
Last edited by a moderator:
Calculate the theoretical % hydrolysis for 1 M solution of Na2CO3

Homework Statement


Calculate the theoretical % hydrolysis for 1 M solution of Na2CO3


Homework Equations


CO3^(2-) + 2H2O -> H2CO3 + 2OH^(-1)

or

CO3^(2-) + H2O -> HCO3^(-1) + OH^(-1)
HCO3^(-1) + H2O -> H2CO3 + OH^(-1)

Chart values for the Ka1 and Ka2 of H2CO3:
Ka1 = 4.3x10^(-7)
Ka2 = 5.6x10^(-11)

The Attempt at a Solution



Kb1 = {[HCO3^(-1)][OH^(-1)]}/[CO3^(-2)] = Kw/Ka2 = [10^(-14)]/[5.6x10^(-11) = 1.78x10^(-4)

Kb2 = {[H2CO3][OH^(-1)]}/[HCO3^(-1)] = Kw/Ka1 = [10^(-14)]/[4.3x10^(-7)] = 2.32x10^(-8)
 
Moderator's note: merged two threads.
 


imconfused said:
CO3^(2-) + H2O -> HCO3^(-1) + OH^(-1)

You know equilibrium concentration of OH- - can you use it to calculate concentration of HCO3-?

How is hydrolysis percentage defined?
 
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