Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Slinky on May 10, 2005, 02:56:50 PM
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I am having a BLANK understanding the net ionic eqn issue.
"The net ionic equation for the hydrolysis of the salt, Na2S- is, "
...OK so I have up to this point, (i think i have it right)
Na2S(aq) + 2H2O(l)<--> 2NaOH(aq) + H2S
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Na2S(aq) + 2H2O(l)<--> 2NaOH(aq) + H2S
from the overall equation, i break down the aq ionic compounds to their individual ions.
2Na+ + S2- + 2H2O => 2Na+ + 2OH- + H2S
next, i eliminate the common ions on left and right hand side.
S2- + 2H2O => H2S + 2OH-
this final equation is known as the net ionic equation. it shows for the above reaction, effectively it involves the hydrolysis of sulphide anion to form hydrogen sulphide and hydroxide ions. the eliminated sodium ions are spectator ions. They don't actually participate in the reaction although they are present in solution.
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For hydrolysis of Na2S more correct is
S2- + H2O = HS- + OH-
the next step
HS- + H2O = H2S + OH- is negligible (~1%)
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For hydrolysis of Na2S more correct is
S2- + H2O = HS- + OH-
the next step
HS- + H2O = H2S + OH- is negligible (~1%)
AWK, where were you when me and Borek argued about this 2 months ago? ;)
I hope Borek sees this ;)
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AWK, where were you when me and Borek argued about this 2 months ago? ;)
I hope Borek sees this ;)
Yes, I do. And I have no idea what you are referring to. Quoting old thread:
common sense tells me that NaOH will react with H2S. So the equlibrium of that reaction would be moved to the (...) forming of Na2S.
AWK just wrote the opposite. If you dissolve Na2S hydrolysis moves the equlibrium to the formation of NaHS and NaOH and there will be virtually no S2- in the solution. In no way that contradicts what I wrote a month ago. On the contrary - it supports my opinion.
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Look guys, this isn't all that difficult. Na2S will, of course, dissociate in water. Now, what has every chemistry teacher ever said about the anion of a weak acid?! that's right, it acts as a base. This is just a very simple Bronstead-Lowry acid-base equation (remember, a bronstead lowry base is something that accepts a proton, such as ammonia).
S-2 + H20 ====> HS- + OH-
Here, S-2 is a base (accepts a proton), with it's conjugate acid being HS-; H20 acts as an acid (yes, it can do that, since it donates a proton), with it's conjugate base being OH-.
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H2S has a Ka2 of 1.0 x 10-14 if you are wondering why this can be heated discussion sometimes.
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H2S has a Ka2 of 1.0 x 10-14 if you are wondering why this can be heated discussion sometimes.
Different sources give very different values for pKa2, I have seen values from 12 up to over 20. In any case, that's a strong base - for pKa2 = 14 (value given by geodome) pKb1 = 0, which is comparable with NaOH and KOH (see my page at http://www.chembuddy.com/?left=BATE&right=dissociation_constants and comments at http://www.chembuddy.com/?left=FAQ).