A solution is prepared by adding 300 mL of 0.500 M NH3 and 100 mL of 0.500 M HCl. Assuming that the volumes are additive, what is the pH of the resulting mixture? Kb for ammonia is 1.8 x10 -5

You've added a base to an acid; which one is in excess. The equation is

NH3 + HCl ==> NH4Cl
mmoles NH3 = 300 mL x 0.500 M = 150
mmoles HCl = 100 mL x 0.500 M = 50

It should be obvious that the reaction produces 50 mmoles NH4Cl, all of the HCl is used and there are 100 mmoles NH3 remaining unreacted. This is a solution of a weak base (NH3) and its salt (NH4Cl) and that is a buffered solution. So use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid))]
Post your work if you get stuck.

8.9

To find the pH of the resulting mixture, we need to determine the concentration of the resulting species.

Step 1: Find moles of NH3 and HCl used.
Moles of NH3 = Volume of NH3 (L) × Concentration of NH3 (M)
Moles of NH3 = (300 mL ÷ 1000 mL/L) × 0.500 M
Moles of NH3 = 0.150 moles

Moles of HCl = Volume of HCl (L) × Concentration of HCl (M)
Moles of HCl = (100 mL ÷ 1000 mL/L) × 0.500 M
Moles of HCl = 0.050 moles

Step 2: Determine limiting reactant.
Since HCl and NH3 have a 1:1 stoichiometric ratio, the limiting reactant is the one with fewer moles. In this case, HCl has fewer moles (0.050 moles).

Step 3: Determine moles of excess reactant remaining.
Moles of excess NH3 = Moles of NH3 initially - Moles of NH3 used
Moles of excess NH3 = 0.150 moles - 0.050 moles
Moles of excess NH3 = 0.100 moles

Step 4: Determine concentration of excess NH3 in the resulting mixture.
Volume of resulting mixture = 300 mL + 100 mL = 400 mL
Concentration of excess NH3 in the resulting mixture = Moles of excess NH3 / Volume of resulting mixture
Concentration of excess NH3 = (0.100 moles / 400 mL) × (1000 mL/L)
Concentration of excess NH3 = 0.250 M

Step 5: Determine concentration of NH4+ in the resulting mixture.
Since NH3 reacted with HCl in a 1:1 ratio, the concentration of NH4+ in the resulting mixture is equal to the concentration of HCl used.
Concentration of NH4+ = 0.500 M

Step 6: Calculate the concentration of OH- ions generated.
Since NH4+ is a weak acid (conjugate acid of NH3), it will undergo hydrolysis:
NH4+ + H2O → NH3 + H3O+

Kw = [H3O+][OH-] = 1.0 × 10^-14

Kw = Ka × Kb
[H3O+][OH-] = (x)(0.500)
1.0 × 10^-14 = (x)(0.500)

x = [H3O+] = (1.0 × 10^-14) / (0.500)
x = 2.0 × 10^-14

Step 7: Calculate the pH of the resulting mixture.
pH = -log[H3O+]
pH = -log(2.0 × 10^-14)
pH ≈ 13.70

Therefore, the pH of the resulting mixture is approximately 13.70.

To find the pH of the resulting mixture, we need to understand the reaction that occurs when NH3 (ammonia) reacts with HCl (hydrochloric acid).

NH3 + HCl → NH4+ + Cl-

In this reaction, NH3 acts as a base and reacts with HCl, which is an acid, to form NH4+ (ammonium) and Cl- (chloride) ions. The NH4+ ion is acidic in nature, so the resulting mixture will be acidic.

To find the pH, we need to determine the concentration of H+ ions (which determines the acidity) in the resulting mixture. We can do this by applying the principles of acid-base equilibria and the Kb value for ammonia.

Given:
Volume of NH3 solution (V1) = 300 mL
Concentration of NH3 solution (C1) = 0.500 M
Volume of HCl solution (V2) = 100 mL
Concentration of HCl solution (C2) = 0.500 M
Kb for ammonia = 1.8 x 10^-5

Step 1: Calculate the number of moles of NH3 and HCl used:
Moles of NH3 (molesNH3) = C1 * V1
= 0.500 M * 0.3 L (converting mL to L)
= 0.150 moles

Moles of HCl (molesHCl) = C2 * V2
= 0.500 M * 0.1 L (converting mL to L)
= 0.050 moles

Step 2: Determine the limiting reactant:
The reactant that is completely consumed and determines the amount of product formed is the limiting reactant. In this case, NH3 and HCl react in a 1:1 ratio, so molesNH3 = molesHCl. Therefore, neither reactant is limiting.

Step 3: Calculate the concentration of NH4+ ions in the resulting mixture:
Since molesNH3 = molesHCl, the amount of NH4+ ions formed is equal to the moles of HCl used.

Concentration of NH4+ ions = (molesHCl) / (total volume of the resulting mixture)

Total volume of the resulting mixture = V1 + V2
= 0.3 L + 0.1 L
= 0.4 L

Concentration of NH4+ ions = (0.050 moles) / (0.4 L)
= 0.125 M

Step 4: Calculate the concentration of OH- ions in the resulting mixture:
Since NH4+ ions are formed through the reaction of NH3 with HCl, there will also be a corresponding production of OH- ions due to the base property of NH3.

Using the Kb expression for ammonia:
Kb = [NH4+][OH-] / [NH3]

Plugging in the known values:
1.8 x 10^-5 = (0.125 M)(x) / (0.150 M)

Solving for x (concentration of OH- ions):
x = (1.8 x 10^-5)(0.150 M) / (0.125 M)
= 2.16 x 10^-5 M

Step 5: Calculate the concentration of H+ ions in the resulting mixture:
To find the concentration of H+ ions, we need to use the equation Kw = [H+][OH-]. At 25°C, Kw = 1.0 x 10^-14.

[H+][OH-] = 1.0 x 10^-14

[H+](2.16 x 10^-5) = 1.0 x 10^-14

[H+] = (1.0 x 10^-14) / (2.16 x 10^-5)
= 4.63 x 10^-10 M

Step 6: Calculate the pH of the resulting mixture:
pH = -log[H+]
= -log(4.63 x 10^-10)
= 9.34

Therefore, the pH of the resulting mixture is approximately 9.34.