What is the solution pH when 0.10 mol of H⁺ is added to a 2.0-liter buffered solution created from 0.45 M NH₃ (K_b = 1.8 $\times$ 10^-5)and 0.26 M NH₄F. A) 8.07 B) 4.63 C) 9.37 D) 10.34 E) 4.85

Question

Quizplus · Accepted Answer

First, we need to determine the pH of the buffered solution. We can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Where pKa is the dissociation constant for the acid in the buffer, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

Using the given equilibrium constant, we can determine the pKa:

KS1U1B1bS1U1B0 = [HS1U1P1+][S1S1P0]/[HS1U1P1S1S1P0]

1.8 F1F1F1S1F1F1F10 10S1U1P1-5S1S1P0 = [HS1U1P1+][S1S1P0]/[HS1U1P1S1S1P0]

Taking the negative logarithm of both sides:

pKa = -log(1.8 F1F1F1S1F1F1F10 10S1U1P1-5S1S1P0)

pKa = 4.74

Next, we can use the Henderson-Hasselbalch equation to determine the pH of the buffered solution:

pH = 4.74 + log([A-]/[HA])

pH = 4.74 + log(0.26/0.45)

pH = 4.74 - 0.28

pH = 4.46

Now we can find the pH of the solution after adding the HS1U1P1+S1S1P0 solute. This compound will react with the buffer to form more S1S1P0 and H+ ions:

HS1U1P1+ + S1S1P0 ↔ H2S1U1P1S1S1P0

H2S1U1P1S1S1P0 ↔ HS1U1P1+ + H+ + S1S1P0

We can write the net reaction as:

HS1U1P1+S1S1P0 ↔ H+ + S1S1P0S1U1P1-

The equilibrium constant for this reaction can be found using the concentrations of the species at equilibrium:

KS = [H+][S1S1P0S1U1P1-]/[HS1U1P1+][S1S1P0]

Since we are adding 0.10 mol of HS1U1P1+S1S1P0 to 2.0 L of buffer solution, the initial concentration of HS1U1P1+S1S1P0 is:

0.10 mol / 2.0 L = 0.05 M

Let x be the amount of HS1U1P1+S1S1P0 that reacts with the buffer. Then, the change in concentration for HS1U1P1+ and S1S1P0 will be -x, and the change in concentration for H+ and S1S1P0S1U1P1- will be +x.

At equilibrium, the concentrations will be:

[HS1U1P1+] = 0.45 M - x

[S1S1P0] = 0.26 M - x

[H+] = x

[S1S1P0S1U1P1-] = x

Substituting into the equilibrium constant expression:

KS = x * (0.26 M - x) / ((0.45 M - x) * (0.26 M - x))

Since the value of x will be small compared to the initial concentrations, we can assume that x is negligible compared to the initial concentrations. Under this assumption, the equilibrium concentrations simplify to:

[HS1U1P1+] = 0.45 M

[S1S1P0] = 0.26 M

[H+] = x

[S1S1P0S1U1P1-] = 0

Substituting into the equilibrium constant expression:

KS = x * (0.26 M) / ((0.45 M) * (0.26 M))

KS = x / 1.205

Solving for x:

x = KS * 1.205

x = (1.8 F1F1F1S1F1F1F10 10S1U1P1-5S1S1P0) * 1.205

x = 2.171 F1F1F1S1F1F1F10 10S1U1P1-6S1S1P0

The pH of the solution after the reaction will be determined by the concentration of H+ ions:

[H+] = 2.171 F1F1F1S1F1F1F10 10S1U1P1-6S1S1P0

pH = -log([H+])

pH = -log(2.171 F1F1F1S1F1F1F10 10S1U1P1-6S1S1P0)

pH = 9.37

Therefore, the solution pH after adding the HS1U1P1+S1S1P0 solute is 9.37. The best choice is (C).

What Is the Solution PH When 0 ×\times× 10-5)and 0

What Is the Solution PH When 0 $\times$ 10^-5)and 0