Question

Ammonia (NH₂) is a weak base with a K, value of 1.77 x 105. Hydrolodic acid (HI) is a strong acid with a K₂ value of 3.2 x 109

32.00-mL of a 1.50 M NH3 solution is being titrated with a 1.75 M HI solution.
What is the pH when 11.00 mL of the 1.75 MHI solution has been added?
Hint: Remember that this is a base equilibrium type of problem so you have OH(aq) as a product and not H₂0" (04)
O 9.27
O 9.80
O 9.01
O 10.04
O 9.42
O 8.93
O 9.68
O 9.35
O 9.92
O 10.27
09.56
O 10.16
O 9.19
O 8.85
NH3(0)
+₂0)
Nhỏ loa 4 CH

Answer 1

To calculate the pH when 11.00 mL of the 1.75 M HI solution has been added to a 32.00-mL of a 1.50 M NH3 solution, we use the neutralization reaction between a weak base (NH3) and a strong acid (HI). From the reaction, we can determine the moles of NH3 remaining and the moles of the resulting salt (NH4I). The concentration of NH4I is calculated, and since it is fully ionized, we can determine the concentration of OH- ions using the Kb value for NH4+. Using the concentration of OH- ions, the pH can be calculated as 14 - pOH. Therefore, the pH is 9.25.

Step-by-step explanation:

To calculate the pH when 11.00 mL of the 1.75 M HI solution has been added to a 32.00-mL of a 1.50 M NH3 solution, we need to determine the moles of each species. First, we calculate the moles of NH3, which is 1.50 M x 32.00 mL = 48.00 mmol. Then, we calculate the moles of HI added, which is 1.75 M x 11.00 mL = 19.25 mmol. Since NH3 is a weak base and HI is a strong acid, the reaction between them will be a neutralization reaction, producing only water and salt.

The moles of NH3 remaining after the reaction is 48.00 mmol - 19.25 mmol = 28.75 mmol. The moles of the resulting salt, NH4I, are equal to the moles of HI added, which is 19.25 mmol. The concentration of NH4I can be calculated as moles/volume = 19.25 mmol / 43.00 mL = 0.45 M. Since NH4I is fully ionized, it will dissociate into NH4+ and I- ions. Since NH4+ is the conjugate acid of NH3, it will hydrolyze, producing OH- ions. To determine the concentration of OH- ions and the resulting pH, we can use the Kb value for NH4+ which is equal to Kw/Ka (water dissociation constant divided by the acid dissociation constant for NH4+).

The Kb value for NH4+ is calculated as 1.0 x 10^-14 / 5.6 x 10^-10 = 1.79 x 10^-5. From this Kb value, we can calculate the concentration of OH- ions using the equation Kb = [OH-][NH4+]/[NH3]. Rearranging the equation, [OH-] = Kb[NH3]/[NH4+] = (1.79 x 10^-5)(0.45)/(0.45) = 1.79 x 10^-5 M. To find the pH, we can use the equation pH = 14 - pOH. Since pOH = -log[OH-], we can calculate pOH as -log(1.79 x 10^-5) = 4.75. Therefore, the pH = 14 - 4.75 = 9.25.