Which Equation Correctly Represents the Reaction Between HCHO2 and K2CO3?

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lkh1986
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The molecular equation for the reaction between HCHO2 and K2CO3 is
2HCHO3 +K2CO3 --> H2CO3 + 2KCHO2.

H2CO3 will undergo further reaction to become H2O and CO2, so maybe the molelar equation can be written as 2HCHO3 +K2CO3 --> H2O + CO2 + 2KCHO2. Which of the molecular equation is the correct one?

Then, for ionic equation, which one of the following is the correct one?
2HCHO2 + 2K+ + CO32- --> H2CO3 + 2K+ + 2CHO2-
or
2HCHO2 + 2K+ + CO32- --> H2O + CO2 + 2K+ + 2CHO2-
or
2H+ + 2CHO2- + 2K+ + CO32- --> H2CO3 + 2K+ + 2CHO2-
or
2H+ + 2CHO2- + 2K+ + CO32- --> H2O + CO2 + 2K+ + 2CHO2-

Do we need to separate the acid? But then again it is a weak acid.
However, since this reaction is a neutralization reaction between an acid and a base, it would be weird if we don't separate the acid into ions. Technically, small percene of the acid do ionize.
 
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lkh1986 said:
The molecular equation for the reaction between HCHO2 and K2CO3 is
2HCHO3 +K2CO3 --> H2CO3 + 2KCHO2.

H2CO3 will undergo further reaction to become H2O and CO2, so maybe the molelar equation can be written as 2HCHO3 +K2CO3 --> H2O + CO2 + 2KCHO2. Which of the molecular equation is the correct one?

The second is correct, because H2CO3 doesn't exist in a solution and always dissociates in H2O and CO2.

lkh1986 said:
Then, for ionic equation, which one of the following is the correct one?
2HCHO2 + 2K+ + CO32- --> H2CO3 + 2K+ + 2CHO2-
or
2HCHO2 + 2K+ + CO32- --> H2O + CO2 + 2K+ + 2CHO2-
or
2H+ + 2CHO2- + 2K+ + CO32- --> H2CO3 + 2K+ + 2CHO2-
or
2H+ + 2CHO2- + 2K+ + CO32- --> H2O + CO2 + 2K+ + 2CHO2-

none are correct, since you have equal amount K+ on both sides of the arrow so they cancel each other out. But apart from that i would go for the second one.

lkh1986 said:
Do we need to separate the acid? But then again it is a weak acid.
However, since this reaction is a neutralization reaction between an acid and a base, it would be weird if we don't separate the acid into ions. Technically, small percene of the acid do ionize.

assume the reaction takes place completely all HCOOH will form HCOO-