Question

Is NaCHO2 strong base? Please explain why as well.

Answer 1

No, NaCHO2 is a weak base.

Ans. Reaction 1:        HCOOH + H₂O -----------> HCOO- + H₃O⁺

            Ka = [HCOO^-] [H3O⁺] / [HCOOH]              - equation 1

            Or, [HCOOH] = [HCOO^-] [H3O⁺]/ Ka         - equation 2

Reaction 2:                 HCOO^- + H₂O ---------> HCOOH + OH^-

            Kb = [HCOOH] [OH^-] / [HCOO^-]                 - equation 3

Putting the value of equation [HCOOH] from equation 2 in equation 3-

            Kb = ([HCOO^-] [H3O⁺]/ Ka) [OH^-] / [HCOO^-]

            Ok, Kb x Ka = [H3O⁺] [OH^-] = 10^-14                       ; [Note: [H3O⁺] [OH^-] = 10^-14]

            Or, Kb = 10^-14 / Ka = 10^-14 / (1.8 x 10^-4) = 5.6 x 10^-11

Note that Kb of formate ion is exceedingly small. The Kb value of strong bases ranges in several orders of 10. For example, the Kb of NaOH is around 10²⁰.

Therefore, formate ion (or its sodium salt) is NOT a strong base.

NaCHO₂ is a weak base as its Kb is around 5.6 x 10^-11. Smaller is the Kb value, weaker is the base.

Note: 1. Weaker is the acid, stronger base is its conjugate base.

The above statement is used to predict the relative strength of conjugate bases derived from a set of weak acids. However, these conjugate bases are weak base under normal conditions for the weak acids used in general.

Note 2: The above statement also follows the formula- Ka x Kb = 10^-14.

Since Ka value of formic acid is low (but NOT exceedingly low, say in order of 10^-23), the resultant Kb of conjugate base is also very low.

If the Ka of a hypothetical acid, say AH is 10^-23 , the Kb of its conjugate base would be Kb = 10^-14 / 10^-23 = 10⁹ . Thus, the conjugate base has Kb of 10⁹ and wopuld be a strong base.

However, our conjugate base, HCOO- is a weak base.

Ans. Reaction 1:        HCOOH + H₂O -----------> HCOO- + H₃O⁺

            Ka = [HCOO^-] [H3O⁺] / [HCOOH]              - equation 1

            Or, [HCOOH] = [HCOO^-] [H3O⁺]/ Ka         - equation 2

Reaction 2:                 HCOO^- + H₂O ---------> HCOOH + OH^-

            Kb = [HCOOH] [OH^-] / [HCOO^-]                 - equation 3

Putting the value of equation [HCOOH] from equation 2 in equation 3-

            Kb = ([HCOO^-] [H3O⁺]/ Ka) [OH^-] / [HCOO^-]

            Ok, Kb x Ka = [H3O⁺] [OH^-] = 10^-14                       ; [Note: [H3O⁺] [OH^-] = 10^-14]

            Or, Kb = 10^-14 / Ka = 10^-14 / (1.8 x 10^-4) = 5.6 x 10^-11

Note that Kb of formate ion is exceedingly small. The Kb value of strong bases ranges in several orders of 10. For example, the Kb of NaOH is around 10²⁰.

Therefore, formate ion (or its sodium salt) is NOT a strong base.

NaCHO₂ is a weak base as its Kb is around 5.6 x 10^-11. Smaller is the Kb value, weaker is the base.

Note: 1. Weaker is the acid, stronger base is its conjugate base.

The above statement is used to predict the relative strength of conjugate bases derived from a set of weak acids. However, these conjugate bases are weak base under normal conditions for the weak acids used in general.

Note 2: The above statement also follows the formula- Ka x Kb = 10^-14.

Since Ka value of formic acid is low (but NOT exceedingly low, say in order of 10^-23), the resultant Kb of conjugate base is also very low.

If the Ka of a hypothetical acid, say AH is 10^-23 , the Kb of its conjugate base would be Kb = 10^-14 / 10^-23 = 10⁹ . Thus, the conjugate base has Kb of 10⁹ and wopuld be a strong base.

However, our conjugate base, HCOO- is a weak base.