Answer:
(a) Hypochlorous acid (HOCl)
(b) Formic acid (HCOOH)
(c) Product side
(d) Greater than 1
Explanation:
<u>The chemical reaction is</u>: HCOOH + NaOCl ⇌ HCOONa + HOCl
In this reaction, formic acid (HCOOH) reacts with sodium hypochlorite (NaOCl) to give sodium formate (HCOONa) and hypochlorous acid (HOCl).
(a) In this reaction, <u>hypochlorous acid (HOCl) is a conjugate acid of the base, hypochlorite ion (ClO⁻).</u>
<u>Therefore, the acid on the product side is hypochlorous acid (HOCl).</u>
(b) Given: KaHCOOH = 2 × 10⁻⁴, KaHOCl = 4 × 10⁻⁸
Acid dissociation constant (Ka) of a given acid gives its acid strength and is equal to the equilibrium constant of a given acid dissociation reaction.
Since, Ka of formic acid > Ka of hypochlorous acid
<u>Therefore, formic acid is a stronger acid than hypochlorous acid.</u>
(c) KaHCOOH = [HCOONa] [H] / [HCOOH]= 2 × 10⁻⁴
KaHOCl = [NaOCl] [H] / [HOCl] = 4 × 10⁻⁸
As the acid dissociation constant of formic acid is greater than hypochlorous acid. So, the <u>dissociation of formic acid (HCOOH) to give the product sodium formate (HCOONa) is favored.</u>
<u>Therefore, the forward reaction or the product side of the reaction is favored.</u>
(d) The equilibrium constant: K = [HCOONa] [HOCl] / [HCOOH] [NaOCl]
As the forward reaction or the formation of the products is favored. <u>Therefore the value of equilibrium constant for the given reaction > 1.</u>