Upping Hardness

[aqua_chem]

Does it lower ph at high ph levels? As per chart towards bottom of link. Kind of confused here.

I'm guessing the HC03(-) goes to ---> H(+) + C03(--) but don't really understand why if you wouldn't mind explaining? Or any links (need to be simple though as haven't done chemistry in a very long time). Any help here much appreciated.

Carbon Dioxide and Carbonic Acid

Ask and you shall receive!


It's all because of these equation from that site:

H2CO3 (aq) <-> H+ (aq) + HCO3- (aq)

Equilibrium expression:

turns into:

Because H2CO3 (carbonic acid) rapidly decomposes into CO2 and H2O at room temp and pressure...

And this principle: Le Chatelier's Principle.


If I suddenly add a bunch of bicarbonate (HCO3-), then the system must 'adjust' to make that equation true (ie, Q > K). It does this by reacting with a hydrogen atom, consuming it to make CO2 and H2O, and thus making the system more basic. So there is in the end less HCO3- and less H+, more CO2 is made, and nobody cares about water. The equation is then balanced again (Q = K).

The reaction in the other direction:

HCO3- (aq) <-> H+ (aq) + CO32- (aq)

doesn't happen much at pH below about 8 because CO3-- is a much stronger base than HCO3-, and as such will not yield it's proton until the entire solution is much more basic.

BIG FAT EDIT: I see your question. I misspoke and will edit my post as such. It increase pH, decreases acidity. You are correct. But I made this big post and deleting it would make me sad.

 

[NigelK8485]

Ask and you shall receive!





It's all because of these equation from that site:



H2CO3 (aq) <-> H+ (aq) + HCO3- (aq)



Equilibrium expression:

turns into:

Because H2CO3 (carbonic acid) rapidly decomposes into CO2 and H2O at room temp and pressure...



And this principle: Le Chatelier's Principle.





If I suddenly add a bunch of bicarbonate (HCO3-), then the system must 'adjust' to make that equation true (ie, Q > K). It does this by reacting with a hydrogen atom, consuming it to make CO2 and H2O, and thus making the system more basic. So there is in the end less HCO3- and less H+, more CO2 is made, and nobody cares about water. The equation is then balanced again (Q = K).



The reaction in the other direction:



HCO3- (aq) <-> H+ (aq) + CO32- (aq)



doesn't happen much at pH below about 8 because CO3-- is a much stronger base than HCO3-, and as such will not yield it's proton until the entire solution is much more basic.



BIG FAT EDIT: I see your question. I misspoke and will edit my post as such. It increase pH, decreases acidity. You are correct. But I made this big post and deleting it would make me sad.

It's a very impressive post, even if it didn't answer the question!


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[DeeLee2013]

Long live SCIENCE!!


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[Delapool]

Ask and you shall receive!





It's all because of these equation from that site:



H2CO3 (aq) <-> H+ (aq) + HCO3- (aq)



Equilibrium expression:

turns into:

Because H2CO3 (carbonic acid) rapidly decomposes into CO2 and H2O at room temp and pressure...



And this principle: Le Chatelier's Principle.





If I suddenly add a bunch of bicarbonate (HCO3-), then the system must 'adjust' to make that equation true (ie, Q > K). It does this by reacting with a hydrogen atom, consuming it to make CO2 and H2O, and thus making the system more basic. So there is in the end less HCO3- and less H+, more CO2 is made, and nobody cares about water. The equation is then balanced again (Q = K).



The reaction in the other direction:



HCO3- (aq) <-> H+ (aq) + CO32- (aq)



doesn't happen much at pH below about 8 because CO3-- is a much stronger base than HCO3-, and as such will not yield it's proton until the entire solution is much more basic.



BIG FAT EDIT: I see your question. I misspoke and will edit my post as such. It increase pH, decreases acidity. You are correct. But I made this big post and deleting it would make me sad.

Hi, many many thanks for this. Helps a lot. Will re-read again but this adds to the picture. Thanks again.