MadSci Network: Chemistry

Re: Why does baking soda dissolves better in water than salt does?

Date: Thu Apr 23 20:35:35 1998
Posted By: Michael Weibel, Grad student Chemistry/Physics, University of Utah
Area of science: Chemistry
ID: 889551197.Ch

Hi Ken.  According to general solubility rules, you can predict that all 
alkali metal compounds are soluble in water (see any general chemistry text 
for a list of general solubility rules).  This means that compounds 
containing lithium (Li), sodium (Na), potassium (K), rubidium (Rb) and 
cesium (Cs) are soluble in water.  Salt (sodium chloride aka NaCl) and 
baking soda (aka sodium bicarbonate aka NaHCO3) both fall into this 
category, and should dissolve in water.

Now, take some water and try to dissolve some measured quantity of each 
substance in it.  According to the 74th edition of the CRC Handbook of 
Chemistry and Physics, at zero degrees centigrade (ice water), you can 
dissolve 6.9 grams of sodium bicarbonate in 100 mL of water (about 2/7 of 
the volume in a can of soda), but 35.7 grams of salt should dissolve in the 
same (separate) container of water.  At 100 degrees centigrade (the 
boiling point of water at sea level...the boiling point of water drops a 
few degrees if you are at a high elevation like in Salt Lake City...approx 
4500 feet above sea level), 39.1 grams of sodium chloride dissolve in 100 
mL of water, so between these two temperatures, the solubility of salt in 
water isn't changing very much.  At 60 degrees centigrade, only 16.4 grams 
of sodium bicarbonate dissolves in 100 mL of water.  It would appear that 
salt is MORE soluble than baking soda over these temperatures.

I'd guess that the salt is more soluble than baking soda because water is 
polar (the oxygen part of water can be thought of as being more negatively 
charged than the hydrogen parts) and salt is more polar than sodium 
bicarbonate.  You might have heard the saying "like dissolves like"...very 
polar things dissolve best in very polar liquids.

I hope this answered your question.
Please feel free to email me if you have any further questions.

Best Regards,
Mike Weibel

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