Aqueous solutions of formic acid, hcooh, and sodium hypochlorite are mixed and a bronsted acid/base reaction occurs. formic acid: kahcooh = 2 x 10-4 hypochlorous acid: kahocl = 4 x 10-8 in the net equation: a) what is the name of the acid on the product side? type in the name. acid b) which of the two acids is stronger? type in the name. acid c) the favored side is (product/reactant) d) the equilibrium constant is (greater/less) than 1

(a) Hypochlorous acid (HOCl)

(b) Formic acid (HCOOH)

(c) Product side

(d) Greater than 1

Explanation:

The chemical reaction is: 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, hypochlorous acid (HOCl) is a conjugate acid of the base, hypochlorite ion (ClO⁻).

Therefore, the acid on the product side is hypochlorous acid (HOCl).

(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

Therefore, formic acid is a stronger acid than hypochlorous acid.

(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 dissociation of formic acid (HCOOH) to give the product sodium formate (HCOONa) is favored.

Therefore, the forward reaction or the product side of the reaction is favored.

(d) The equilibrium constant: K = [HCOONa] [HOCl] / [HCOOH] [NaOCl]

As the forward reaction or the formation of the products is favored. Therefore the value of equilibrium constant for the given reaction > 1.

the answer is b & c i believe : )

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )