A solution is prepared by mixing 100 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 1.8x10-5
To find the pH of the resulting mixture, we need to calculate the concentration of OH- ions first. The reaction between NH3 (ammonia) and HCl (hydrochloric acid) will produce NH4+ (ammonium) and Cl- ions.
First, let's calculate the moles of NH3 and HCl used in the mixture:
Moles of NH3 = (0.500 M) x (100 mL / 1000 mL) = 0.050 mol
Moles of HCl = (0.500 M) x (100 mL / 1000 mL) = 0.050 mol
Since the volumes are additive, the total volume of the resulting mixture is 100 mL + 100 mL = 200 mL.
Next, let's calculate the concentration of NH4+ produced:
Moles of NH4+ = Moles of NH3 = 0.050 mol
Concentration of NH4+ = Moles of NH4+ / Total volume = 0.050 mol / 200 mL = 0.250 M
Now, let's calculate the OH- concentration using the Kb value:
Kb = [NH4+][OH-] / [NH3]
[OH-] = Kb x [NH3] / [NH4+]
= (1.8x10^-5) x (0.250 M) / (0.500 M)
= 9.0x10^-6 M
Finally, let's calculate the pOH and then the pH:
pOH = -log10[OH-] = -log10(9.0x10^-6) = 5.05
pH = 14 - pOH = 14 - 5.05 = 8.95
Therefore, the pH of the resulting mixture is approximately 8.95.
To find the pH of the resulting mixture, we need to calculate the concentration of the remaining species and determine if the solution is acidic or basic.
First, let's calculate the amount of NH3 and HCl in moles that exist in the solution.
Molarity (M) = Moles (mol) / Volume (L)
Using the given volumes and molarities:
For NH3:
Molarity of NH3 = 0.500 M
Volume of NH3 = 100 mL = 0.100 L
Moles of NH3 = Molarity × Volume
= 0.500 M × 0.100 L
= 0.050 mol
For HCl:
Molarity of HCl = 0.500 M
Volume of HCl = 100 mL = 0.100 L
Moles of HCl = Molarity × Volume
= 0.500 M × 0.100 L
= 0.050 mol
Since HCl and NH3 react stoichiometrically in a 1:1 ratio, the amount of HCl and NH3 present after the reaction is equal, and there is no excess of either.
Now, let's consider the reaction between NH3 and HCl:
NH3 + HCl → NH4+ + Cl-
The NH3 reacts with HCl to form NH4+ (ammonium) and Cl- (chloride) ions.
The NH4+ ions can react with water to form NH3 and H3O+ (hydronium) ions:
NH4+ + H2O → NH3 + H3O+
Since NH4+ is a weak acid, it will partially dissociate to form NH3 and H3O+.
Now, let's calculate the concentration of NH3 and H3O+ ions in the solution.
First, calculate the concentration of NH3:
Concentration of NH3 (initial) = Moles of NH3 / Volume of solution
Moles of NH3 = Moles of NH3 remaining after reaction = 0.050 mol
Volume of solution = Volume of NH3 + Volume of HCl = 0.100 L + 0.100 L = 0.200 L
Concentration of NH3 = 0.050 mol / 0.200 L
= 0.250 M
Next, calculate the concentration of H3O+ ions:
Since NH3 acts as a weak base, it undergoes hydrolysis with water to produce OH- (hydroxide) ions. However, since Kb for NH3 is given, we can use it to calculate the concentration of OH- and then convert it to the concentration of H3O+ using the relation Kw = [H3O+][OH-] = 1.0 × 10^-14.
Kb = [NH4+][OH-] / [NH3]
With the given Kb = 1.8 × 10^-5, we can calculate [OH-] using the concentration of NH3:
[NH4+] = [OH-] (from the reaction)
[NH4+] = [NH3 (initial)] - [NH3 (remaining)]
[NH4+] = 0.500 M - 0.250 M
[NH4+] = 0.250 M
Substituting the values into the Kb expression:
1.8 × 10^-5 = (0.250 M)([OH-]) / (0.250 M)
Simplifying, [OH-] = 1.8 × 10^-5 M
Using Kw = [H3O+][OH-] = 1.0 × 10^-14,
[H3O+] = 1.0 × 10^-14 / [OH-]
[H3O+] = 1.0 × 10^-14 / (1.8 × 10^-5)
[H3O+] ≈ 5.56 × 10^-10 M
Since the solution is basic due to the presence of OH- ions, the pH can be calculated using the relation:
pH = -log10[H3O+]
pH = -log10(5.56 × 10^-10)
pOH = -log10[OH-] = -log10(1.8 × 10^-5)
Calculating the pOH:
pOH = -log10(1.8 × 10^-5)
pOH ≈ 4.744
pH = 14 - pOH
pH ≈ 14 - 4.744
pH ≈ 9.256
Therefore, the pH of the resulting mixture is approximately 9.256.