1) Find the pH of 0.54 M C5H5N(aq) given that its Kb = 1.7×10^-9.

The answer I got is 2.75 but I got it wrong. The answer is supposed to be 9.48. Why?

2) Find the pH of a 0.864 M NH4Cl solution given that the Kb of NH3 = 1.8x10^-5 at 25 °C.

How would I start off this?

1) Find the pH of 0.54 M C5H5N(aq) given that its Kb = 1.7×10^-9.

The answer I got is 2.75 but I got it wrong. The answer is supposed to be 9.48. Why?

2) Find the pH of a 0.864 M NH4Cl solution given that the Kb of NH3 = 1.8x10^-5 at 25 °C.

How would I start off this?

1) To solve this problem, we need to use the relationship between the equilibrium constant (Kb) and the concentration of the conjugate acid-base pair.

C5H5N is a weak base, which means it can accept a proton (H+) to form its conjugate acid, C5H6N+.

The equilibrium reaction is: C5H5N + H2O ⇌ C5H6N+ + OH-

The Kb expression for this reaction is: Kb = [C5H6N+][OH-] / [C5H5N]

Given that Kb = 1.7×10^-9, we can assume that the concentration of OH- ions produced is much smaller compared to the concentration of C5H5N.

To calculate the pH, we need to find the concentration of OH- ions at equilibrium. We can assume that OH- concentration is x (small compared to C5H5N concentration).

Since C5H5N dissociates completely into C5H6N+ and OH- ions, the concentration of OH- ions will be equal to x.

Using the Kb expression, we have: Kb = x * x / (0.54 - x)

Since x is much smaller compared to 0.54, we can approximate 0.54 - x to be 0.54.

Thus, Kb = x * x / 0.54

Simplifying, Kb = x^2 / 0.54

Now, substitute the given Kb value and solve for x:

1.7×10^-9 = x^2 / 0.54

Rearranging the equation gives: x^2 = 1.7×10^-9 * 0.54

Taking the square root of both sides gives: x = √(1.7×10^-9 * 0.54)

Evaluating this expression gives x ≈ 3.19×10^-5

Now, the concentration of OH- ions is approximately 3.19×10^-5 M.

To find the pH, we can use the fact that the concentration of H+ ions (from the dissociation of water) is equal to the concentration of OH- ions at 25 °C, which is 3.19×10^-5 M.

The pH can be calculated using the equation: pH = -log[H+]

pH = -log(3.19×10^-5) ≈ 4.49 (not 2.75)

Hence, the corrected pH value is approximately 4.49, not 2.75.

2) To find the pH of the NH4Cl solution, we need to determine the concentration of H+ ions in the solution, which will determine the acidity or basicity of the solution.

NH4Cl is a salt that dissociates in water to form NH4+ and Cl- ions. NH4+ acts as a weak acid and can donate a proton (H+) in water.

The equilibrium reaction is: NH4+ + H2O ⇌ NH3 + H3O+

To find the concentration of H+ ions, we need to find the concentration of NH4+ ions at equilibrium. We can assume that NH4+ concentration is x (small compared to the initial concentration of NH4Cl).

Since NH4Cl dissociates completely into NH4+ and Cl- ions, the concentration of NH4+ ions will be equal to x.

Using the dissociation expression for NH4+, we have: [NH4+] = x

Since x is much smaller compared to 0.864 (initial concentration of NH4Cl), we can neglect its effect.

Therefore, the concentration of H+ ions in the solution is approximately equal to x.

Given that the Kb of NH3 is 1.8x10^-5, we can assume that the concentration of NH3 (from the dissociation of NH4+) is much smaller compared to the initial concentration of NH4Cl.

Therefore, we can write the Kb expression for NH3 as: Kb = [NH3][H3O+] / [NH4+]

Since [NH3] is much smaller compared to [NH4+], we can neglect its effect.

Thus, Kb = [H3O+]/[NH4+]

To solve for [H3O+] (concentration of H+ ions), we can rearrange the equation:

[H3O+] = Kb * [NH4+]

Substituting the given Kb value and [NH4+] (which is equal to x):

[H3O+] = (1.8x10^-5)(x)

Now, substitute the concentration of H+ ions ([H3O+]) into the pH equation: pH = -log[H+]

pH = -log(1.8x10^-5)(x)

To determine x, we can use the fact that NH4Cl completely dissociates into NH4+ and Cl- ions.

Therefore, the initial concentration of NH4+ is equal to the initial concentration of NH4Cl, which is 0.864 M.

Substituting 0.864 M for x gives:

pH = -log(1.8x10^-5)(0.864)

Simplifying this equation will give you the final answer for the pH of the NH4Cl solution.

1) To find the pH of a solution of a weak base like C5H5N, which is also known as pyridine, we need to use the equilibrium expression for the ionization of the base in water. The balanced equation for the ionization of pyridine is as follows:

C5H5N(aq) + H2O(l) ⇌ C5H5NH+(aq) + OH-(aq)

The equilibrium constant for this reaction is the base ionization constant, Kb. For pyridine, Kb is given as 1.7×10^-9.

To solve for the pH, we first need to determine the concentration of the hydroxide ion (OH-) in the solution at equilibrium. It is important to note that in this case, the concentration of OH- is not equal to the concentration of the base (C5H5N). The concentration of OH- can be determined using the equation:

Kb = [C5H5NH+][OH-] / [C5H5N]

Since the initial concentration of the base (C5H5N) is 0.54 M, we can assume that at equilibrium, the concentration of OH- is also x M. Thus, the concentration of C5H5NH+ will also be x M.

Substituting these into the equilibrium expression, we get:

1.7×10^-9 = (x)(x) / (0.54 - x)

At this point, we can solve this equation using the quadratic formula or make an approximation since x is expected to be very small compared to 0.54. Assuming that x is negligible, we can approximate it to be approximately 0 in the denominator, so the equation becomes:

1.7×10^-9 = x^2 / 0.54

To solve for x, we multiply both sides of the equation by 0.54:

1.7×10^-9 × 0.54 = x^2

x^2 = 9.18×10^-10

Taking the square root of both sides:

x ≈ 3.03×10^-5

Since we assumed x to be negligible compared to 0.54, our approximation was valid. Now we can determine the pOH of the solution using the concentration of OH-, which is x:

pOH = -log10(x) = -log10(3.03×10^-5) ≈ 4.52

Finally, we can calculate the pH using the equation:

pH = 14 - pOH = 14 - 4.52 ≈ 9.48

Therefore, the correct answer is indeed pH ≈ 9.48.

2) To find the pH of the NH4Cl solution, we first need to recognize that NH4Cl is a salt. When dissolved in water, it dissociates into NH4+ and Cl- ions.

The NH4+ ion is the conjugate acid of NH3 (ammonia), which is a weak base. NH3 can accept a proton and act as a base in solution. The equilibrium equation for the reaction of NH3 with water is:

NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)

The Kb value for NH3 is given as 1.8x10^-5.

Since NH4Cl is a salt, it only provides NH4+ ions in solution. Hence, we can think of this as an acidic solution. To find the pH, we need to determine the concentration of the hydronium ion (H3O+) in the solution at equilibrium.

From the equilibrium equation, we can see that for every NH4+ ion formed, one OH- ion is formed as well. Therefore, the concentration of OH- is equal to the concentration of NH4+.

Let's assume that at equilibrium, the concentration of OH- and NH4+ is x M. Since the initial concentration of the NH4Cl solution is 0.864 M, we can assume that the concentration of NH4+ is also x M.

Using the equilibrium expression for the reaction with Kb:

Kb = [NH4+][OH-] / [NH3]

Plugging the concentrations into the equation:

1.8x10^-5 = (x)(x) / (0.864 - x)

As in the previous example, we can make an approximation since x is expected to be very small compared to 0.864. Assuming that x is negligible, we can approximate it to be approximately 0 in the denominator:

1.8x10^-5 ≈ x^2 / 0.864

To solve for x, we multiply both sides of the equation by 0.864:

1.8x10^-5 × 0.864 ≈ x^2

x^2 ≈ 1.5552x10^-5

Taking the square root of both sides:

x ≈ 3.94x10^-3

Since we assumed x to be negligible compared to 0.864, our approximation was valid. Now we can determine the pOH of the solution using the concentration of OH-, which is x:

pOH = -log10(x) = -log10(3.94x10^-3) ≈ 2.4

Finally, we can calculate the pH using the equation:

pH = 14 - pOH = 14 - 2.4 ≈ 11.6

Therefore, the pH of the 0.864 M NH4Cl solution is approximately 11.6.