Sure, here are the steps to calculate the pH of a 0.73M solution of ammonium bromide (NH4Br) at 25°C:

1. First, we need to understand that when NH4Br dissolves in water, it dissociates into its ions and reacts with water to form NH3 and H3O+. The reaction is as follows: NH4+ + H2O - NH3 + H3O+

2. The pKb of NH3 is given as 4.75. We can calculate Kb (the base ionization constant) using the formula: Kb = 10^-pKb. So, Kb = 10^-4.75 = 1.78 x 10^-5.

3. Now, we can set up an ICE (Initial, Change, Equilibrium) table to find the concentrations of the reactants and products at equilibrium.

Initial concentrations: [NH4+] = 0.73 M, [NH3] = 0, [H3O+] = 0
Change in concentrations: [NH4+] = -x, [NH3] = +x, [H3O+] = +x
Equilibrium concentrations: [NH4+] = 0.73 - x, [NH3] = x, [H3O+] = x

4. We can now write the expression for Kb: Kb = ([NH3][H3O+])/[NH4+]. Substituting the equilibrium concentrations into this expression gives: 1.78 x 10^-5 = (x * x) / (0.73 - x).

5. Assuming x is small compared to 0.73, the equation simplifies to: 1.78 x 10^-5 = x^2 / 0.73. Solving for x gives: x = sqrt(1.78 x 10^-5 * 0.73) = 0.0036 M.

6. The concentration of H3O+ (x) is equal to the hydronium ion concentration, [H3O+]. We can find the pH of the solution using the formula: pH = -log[H3O+]. Substituting the value of [H3O+] into this formula gives: pH = -log(0.0036) = 2.44.

So, the pH of the 0.73M solution of NH4Br at 25°C is approximately 2.44.

Question

Sure, here are the steps to calculate the pH of a 0.73M solution of ammonium bromide (NH4Br) at 25°C:

1. First, we need to understand that when NH4Br dissolves in water, it dissociates into its ions and reacts with water to form NH3 and H3O+. The reaction is as follows: NH4+ + H2O -> NH3 + H3O+

2. The pKb of NH3 is given as 4.75. We can calculate Kb (the base ionization constant) using the formula: Kb = 10^-pKb. So, Kb = 10^-4.75 = 1.78 x 10^-5.

3. Now, we can set up an ICE (Initial, Change, Equilibrium) table to find the concentrations of the reactants and products at equilibrium.

Initial concentrations: [NH4+] = 0.73 M, [NH3] = 0, [H3O+] = 0
   Change in concentrations: [NH4+] = -x, [NH3] = +x, [H3O+] = +x
   Equilibrium concentrations: [NH4+] = 0.73 - x, [NH3] = x, [H3O+] = x

4. We can now write the expression for Kb: Kb = ([NH3][H3O+])/[NH4+]. Substituting the equilibrium concentrations into this expression gives: 1.78 x 10^-5 = (x * x) / (0.73 - x).

5. Assuming x is small compared to 0.73, the equation simplifies to: 1.78 x 10^-5 = x^2 / 0.73. Solving for x gives: x = sqrt(1.78 x 10^-5 * 0.73) = 0.0036 M.

6. The concentration of H3O+ (x) is equal to the hydronium ion concentration, [H3O+]. We can find the pH of the solution using the formula: pH = -log[H3O+]. Substituting the value of [H3O+] into this formula gives: pH = -log(0.0036) = 2.44.

So, the pH of the 0.73M solution of NH4Br at 25°C is approximately 2.44.

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