Question

The pH of 0.040 M hypobromous acid (HOBr) is 5.05. Set up the equilibrium equation for the dissociation of HOBr, and caluculate the value of the acid-dissociation constant. If -X represents the change in HOBr concentration, then the equilibrium concentration of HOBr and H3O+ in terms of X are ________ and _______ respectively. X = _______M. Ka= _______ (2 significant numbers using E to express the power of ten). Please fill in the four blanks and/or BOLD your 4 answers when you reply so I can better understand the answers. Thank you.

Answer 1

The dissociation of HOBr can be written as

Now, the acid dissociation constant K_a is expressed as follows:

Where the concentrations are their equilibrium values.

Given that the starting concentration of HOBr , [HOBr] = 0.040 M

We can create the following ICE table

Initial, M	0.040	0	0
Change, M	-X	+X	+X
Equilibrium, M	0..040-X	X	X

Now, it is already given that the pH of the solution is 5.05.

We know that pH is defined as the negative logarithm of the equilibrium H₃O⁺ concentration.

Hence,

Hence, the value of X is .

Now, from the ICE table, we can write the expression of K_a as follows:

Hence, we can fill the blanks as follows:

If -X represent the change in HOBr concentration, then the equilibrium concentration of HOBr and H₃O⁺ in terms of X are 0.040-X and X respectively.

Note: Answers are rounded to two significant figures.