What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M NH₃ with 10.00 mL of 0.10 M NH₄Cl? Assume that the volume of the solutions are additive and that K_b = 1.8 × 10^-5for NH₃. A) 8.56 B) 9.28 C) 9.95 D) 11.13

Question

Quizplus · Accepted Answer

First, we need to use the equation for calculating the pH of a weak base solution: pH = pKb + log([base]/[acid]), where Kb is the base dissociation constant, [base] is the concentration of the base, and [acid] is the concentration of the conjugate acid.

Since we are given the concentration of the base (NHS1U1B13S1U1B0) and the volume of the solution, we can calculate the moles of base present:

moles of base = (50.00 mL)(0.10 mol/L) = 5.00 × 10S1U1P1-4 mol

Next, we need to calculate the concentration of the conjugate acid (NHS1U1B14S1U1B0Cl) using the balanced chemical equation for the reaction between the two compounds:

NHS1U1B13S1U1B0 + HCl → NHS1U1B14S1U1B0Cl

From the equation, we can see that 1 mole of NHS1U1B13S1U1B0 reacts with 1 mole of HCl to form 1 mole of NHS1U1B14S1U1B0Cl. Therefore, the moles of acid present in the solution are:

moles of acid = (10.00 mL)(0.10 mol/L) = 1.00 ×10S1U1P1-4 mol

Since the volumes of the solutions are additive, the total volume of the solution is:

total volume = 50.00 mL + 10.00 mL = 60.00 mL = 0.060 L

Using the moles of base and acid and the total volume, we can calculate their concentrations:

[base] = moles of base/total volume = 5.00 × 10S1U1P1-4 mol/0.060 L = 8.33 × 10S1U1P1-3 M

[acid] = moles of acid/total volume = 1.00 × 10S1U1P1-4 mol/0.060 L = 1.67 × 10S1U1P1-3 M

Finally, we can plug all of these values into the pH equation:

pH = pKb + log([base]/[acid]) = -log(Kb) + log([base]/[acid])

Since the problem states that KS1U1B1bS1U1B0 = 1.8 × 10S1U1P1-5 S1S1P0for NHS1U1B13S1U1B0, we can calculate Kb as:

Kb = Kw/Ka = 1.0 × 10S1U1P1-14/1.8 × 10S1U1P1-5 = 5.56 × 10S1U1P1-10

Plugging in all the values:

pH = -log(Kb) + log([base]/[acid]) = -log(5.56 × 10S1U1P1-10) + log(8.33 × 10S1U1P1-3/1.67 × 10S1U1P1-3) = 9.95

Therefore, the pH of the solution is 9.95, which corresponds to choice C.

What Is the PH of a Solution Prepared by Mixing