What is the pH of the solution which results from mixing 75 mL of 0.50 M NH₃(aq)and 75 mL of 0.50 HCl(aq)at 25 $\circ$ C? (K_b for NH₃ = 1.8 $\times$ 10^-5)

Question

Quizplus · Accepted Answer

To find the pH of the resulting solution, we need to use the equation for the reaction between NHS1U1B13S1U1B0 and HCl:

NHS1U1B13S1U1B0 + HCl → HNHS1U1B13S1U1B0+ + Cl-

We can see from the equation that the amount of HCl is equal to the amount of NHS1U1B13S1U1B0. Therefore, after mixing, we have a total volume of 150 mL of solution with 0.50 M concentration of NHS1U1B13S1U1B0 and HCl.

Next, we can use the equation for the dissociation of NHS1U1B13S1U1B0:

NHS1U1B13S1U1B0 + H2O → HNHS1U1B13S1U1B0+ + OH-

The equilibrium constant for this reaction is given as KS1U1B1bS1U1B0 = 1.8 F1F1F1S1F1F1F10 10S1U1P1-5S1S1P0. At equilibrium, the concentrations of HNHS1U1B13S1U1B0+ and OH- are the same, and can be calculated as follows:

[H+][NHS1U1B13S1U1B0] = KS1U1B1bS1U1B0 = 1.8 F1F1F1S1F1F1F10 10S1U1P1-5S1S1P0
[H+]2 = KS1U1B1bS1U1B0 / [NHS1U1B13S1U1B0]
[H+]2 = (1.8 F1F1F1S1F1F1F10 10S1U1P1-5S1S1P0) / (0.50 M)
[H+]2 = 3.6 F1F1F1S1F1F1F10 10S1U1P1-4S1S1P0
[H+] = 1.9 F1F1F1S1F1F1F10-2S1S1P0

pH = -log[H+]
pH = -log(1.9 F1F1F1S1F1F1F10-2S1S1P0)
pH = 4.93

Therefore, the pH of the resulting solution is 4.93, which is closest to choice D (4.93)

What Is the PH of the Solution Which Results from Mixing