# So, is baking soda actually a strong base? [duplicate]

Based on my recent thoughts, when baking soda ($\ce{NaHCO3}$) is dissolved in water, the following hydrolysis reaction occurs: $$\ce{NaHCO3 + H2O <=> NaOH + H2CO3}$$ However, $$\ce{H2CO3 -> H2O + CO2 ^}$$ so the actual one is: $$\ce{NaHCO3 -> NaOH + CO2 ^}$$ $\ce{CO2}$ escapes, so the reaction is irreversible, all the $\ce{NaHCO3}$ is converted to $\ce{NaOH}$ and the product is a strong base.

The following reaction is possible: $$\ce{HCO3- <=> H+ + CO3^2-}$$ but it's reversible unless it results in $\ce{H2CO3}$ formation, which is the same as the first one, so it doesn't really matter.

However, I was unable to find any evidence to that but a lot of discussions on whether $\ce{NaHCO3}$ behaves like a weak or a strong base, which put me in doubt. Am I right?

## marked as duplicate by Mithoron, airhuff, Jan, Nilay Ghosh, andselisk♦Nov 2 '17 at 3:41

There is nothing to discuss (unless you find it necessary to compare two known numbers by discussion). $\ce{NaHCO3}$ is a very weak base.
Really, do you have baking soda around? Throw some in the water and see whether $\ce{CO2}$ actually escapes. It doesn't. It will if you heat the solution to near-boiling temperatures; then you'll have $\ce{Na2CO3}$ which is a base of moderate strength. You can't convert it all to $\ce{NaOH}$ this way.