Calculate the H3O+ ion concentration and the pH at the equivalence point when 45.0 mL of 0.4000 M NH3 is mixed with 45.0 mL of 0.4000 M HCl.
Ka=5.6x10-10
To calculate the H3O+ ion concentration and the pH at the equivalence point, we need to determine the moles of NH3 and HCl that react and how they combine to form the equilibrium mixture.
1. Calculate the moles of NH3 and HCl:
Moles of NH3 = volume (in L) x concentration
Moles of NH3 = 0.045 L x 0.4 M = 0.018 moles
Moles of HCl = volume (in L) x concentration
Moles of HCl = 0.045 L x 0.4 M = 0.018 moles
2. NH3 reacts with HCl in a 1:1 stoichiometric ratio to form NH4+ and Cl- ions. At the equivalence point, all moles of NH3 will react with an equal number of moles of HCl, resulting in the formation of NH4+ and Cl- ions.
3. Since the concentration of both NH3 and HCl is the same (0.4 M) and the ratio is 1:1, the concentration of NH4+ and Cl- ions at the equivalence point will also be 0.4 M.
4. To calculate the H3O+ ion concentration, we need to take into account the reaction between NH4+ ions and water.
NH4+ ions react with water to form H3O+ ions:
NH4+ (aq) + H2O (l) ⇌ H3O+ (aq) + NH3 (aq)
The equilibrium expression for this reaction is:
Ka = [H3O+] [NH3] / [NH4+]
Rearranging the equation, we get:
[H3O+] = Ka * ([NH4+] / [NH3])
5. At the equivalence point, as mentioned before, [NH4+] = [Cl-] = 0.4 M.
6. Now, plug the values into the equation to calculate the H3O+ ion concentration:
[H3O+] = (5.6 x 10^-10) * (0.4 M / 0.4 M) = 5.6 x 10^-10 M
7. To calculate the pH, use the equation:
pH = -log[H3O+]
pH = -log(5.6 x 10^-10) ≈ 9.25
Therefore, at the equivalence point, the H3O+ ion concentration is 5.6 x 10^-10 M and the pH is approximately 9.25.