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.

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.