Ammonia and the ammonium ion form a conjugate acid-base pair.

         NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)

Calculate the pH of 0.10 M NH3 if Ka for the NH4+ ion is 5.6 x 10-10

 (a) 4.7 

(b) 5.1

 (c) 5.7 

(d) 8.9

 (e) 11.1
 

To calculate the pH of 0.10 M NH3, we need to consider the equilibrium between NH3 and NH4+.

First, we can write the equation for the equilibrium reaction as follows: NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)

Since NH4+ is the conjugate acid of NH3, we can write the expression for the acid dissociation constant (Ka) as: Ka = [NH4+][OH-]/[NH3]

Given that the Ka for NH4+ is 5.6 x 10-10, we can substitute this value into the equation: 5.6 x 10-10 = [NH4+][OH-]/[NH3]

Since we're interested in the pH, which is a measure of the concentration of H+ ions in a solution, we need to determine the concentration of OH- ions. We can do this by using the relationship between [H+], [OH-], and Kw (the ion product of water): Kw = [H+][OH-] = 1.0 x 10^-14

At 25°C, Kw is known to be 1.0 x 10^-14. Therefore, the concentration of OH- can be calculated by dividing Kw by the concentration of H+ ions: [OH-] = Kw/[H+]

Assuming that [H+] = [OH-], we can substitute this expression into the equation for Ka: 5.6 x 10-10 = [NH4+](Kw/[H+])/[NH3]

Now, let's consider the equilibrium expression for NH3: Ka = [NH4+][OH-]/[NH3]

Since OH- is a product of the reaction and is the only species in the denominator of the expression, we can assume that [OH-] ≈ [NH4+]. Therefore, we can replace [OH-] with [NH4+] in the equation: 5.6 x 10-10 = ([NH4+]^2)/[NH3]

Now, we can substitute the concentration of NH3 into the equation: 5.6 x 10-10 = ([NH4+]^2)/(0.10)

Rearranging the equation, we get: [NH4+]^2 = 5.6 x 10-10 x 0.10

Taking the square root of both sides, we find: [NH4+] = √(5.6 x 10-10 x 0.10)

Calculating this value, we get: [NH4+] ≈ 7.5 x 10^-6

Since we assumed [NH4+] ≈ [OH-] earlier, we can use this value to determine the concentration of OH-: [OH-] ≈ 7.5 x 10^-6

Using the relationship between [OH-] and [H+] (assuming they are equal), we get: [H+] = 1.0 x 10^-14/[OH-] ≈ 1.0 x 10^-14/(7.5 x 10^-6)

Simplifying this expression, we find: [H+] ≈ 1.3 x 10^-9

Finally, we can calculate the pH using the formula: pH = -log[H+]

Plugging in the value for [H+], we get: pH ≈ -log(1.3 x 10^-9)

Calculating this value, we find: pH ≈ 8.9

Therefore, the correct answer is (d) 8.9.

In this problem, we used the concept of equilibrium to calculate the pH of a solution of NH3. We considered the dissociation of NH3 to form NH4+ and OH- ions, and used the acid dissociation constant (Ka) to relate the concentrations of these ions. We also used the relationship between [H+], [OH-], and the ion product of water (Kw) to determine the concentration of OH- ions. Finally, we used the formula for pH (-log[H+]) to calculate the pH of the solution.