the equilibrium constant expression for the ionization of HOBr in water, then calculate

Question

the concentration of HOBr(aq) in an HOBr solution that has [H+] equal to 1.8  10–5 M.

Answer 1

The equilibrium constant expression for the ionization of HOBr in water is:

K = [H+][OBr-]/[HOBr]

We are given that [H+] = 1.8  10–5 M. We need to find the concentration of HOBr(aq), which we can do by rearranging the equilibrium constant expression:

[HOBr] = [H+][OBr-]/K

We need the value of K for this calculation. The ionization of HOBr is an acid-base reaction, and the acid dissociation constant (Ka) can be used as a substitute for K. The expression for Ka is:

Ka = [H+][OBr-]/[HOBr]

The relationship between Ka and K is:

K = Kw/Ka

where Kw is the ion product constant for water, which is 1.0  10-14 at 25°C. Therefore, we can calculate K as:

K = 1.0  10-14 / Ka

The value of Ka for HOBr can be found in a reference source or calculated using the equilibrium concentrations of HOBr, H+, and OBr-. For simplicity, we will assume that the value of Ka for HOBr is 2.3  10-9 at 25°C.

Using this value for Ka, we can calculate K:

K = 1.0  10-14 / 2.3  10-9 = 4.3  10-6

Now we can use the equation we derived earlier to find the concentration of HOBr:

[HOBr] = [H+][OBr-]/K

We are given that [H+] = 1.8  10–5 M. We need to find [OBr-], which we know is equal to [HOBr] because the solution is neutral (the concentrations of H+ and OH- are equal). Therefore:

[HOBr] = [OBr-] = 1.8  10–5 M

[Answer: The concentration of HOBr(aq) in an HOBr solution that has [H+] equal to 1.8  10–5 M is 1.8  10–5 M.]