Final answer:
To calculate the pH of a 0.250 M solution of NH4Br, we can use the dissociation reaction of NH4+ in water. By writing and solving the equilibrium expression, we can determine the concentration of H+ ions and then calculate the pH using the formula -log[H+]. The pH of the solution is approximately 9.25.
Step-by-step explanation:
The pH of a solution can be calculated using the formula:
pH = -log[H+]
In this case, we are given a 0.250 M solution of NH4Br. NH4Br is a salt that dissociates into NH4+ and Br- ions in water. The NH4+ ion can act as an acid by donating a proton, and its conjugate base is NH3. The Ka value for NH4+ is given as 5.6 x 10^-10.
To find the pH of the solution, we need to determine the concentration of H+ ions. Since NH4+ is an acid, it can be considered a source of H+ ions. The dissociation reaction is as follows:
NH4+ + H2O ⇌ NH3 + H3O+
The equilibrium expression for this reaction is:
Ka = [NH3][H3O+]/[NH4+]
We can assume that the concentration of NH3 is the same as the concentration of NH4+ in this solution, since NH4Br is a strong electrolyte and completely dissociates. Therefore, we can write the expression as:
Ka = [NH3][H3O+]/(0.250)
We can rearrange the equation to solve for [H3O+]:
[H3O+] = Ka(0.250)/[NH3]
Substituting the values, we get:
[H3O+] = (5.6 x 10^-10)(0.250)/(0.250) = 5.6 x 10^-10
Finally, we can calculate the pH using the formula:
pH = -log[H3O+]
pH = -log(5.6 x 10^-10)
Simplifying the calculation, we find that the pH of the 0.250 M solution of NH4Br is approximately 9.25.