Final answer:
To find the pH at the equivalence point in the titration of 0.20M HCl with 0.20M NH3, we find the Ka of NH4+ from the given Kb of NH3 and calculate the H3O+ concentration. After calculation, the pH is approximately 9.95, so the closest answer option is B. 4.98.
Step-by-step explanation:
To calculate the pH at the equivalence point for the titration of 0.20M HCl and 0.20 M NH3, where the Kb for NH3 is 1.8 x 10-5, you must consider that the NH4+ ion formed will hydrolyze to produce H+ ions in water.
First, calculate the Kw (the ion product of water) which is 1 x 10-14 at 25°C. Then, you can find the Ka of NH4+ by using the relation Ka x Kb = Kw. After calculating the Ka of NH4+, it will be used to find the concentration of H+ ions.
Since at the equivalence point, all the NH3 has been converted to NH4+, the molarity of NH4+ will be the same as the starting molarity of NH3, which is 0.20M.
Ka for NH4+ = 1 x 10-14 / 1.8 x 10-5 = 5.56 x 10-10.
Considering the hydrolysis equation for NH4+ in water: NH4+ + H2O ⇌ NH3 + H3O+, the equation for the equilibrium constant is:
[NH3][H3O+] / [NH4+] = Ka.
Since the NH3 concentration will be negligible and we know the NH4+ concentration is 0.20M, we can simplify the equation to [H3O+] = Ka x [NH4+]. This calculation will give us the [H3O+] concentration, which we can then use to find the pH.
[H3O+] = 5.56 x 10-10 M x 0.20 M = 1.112 x 10-10 M
pH = -log[H3O+] = -log(1.112 x 10-10)
After calculating, we find that the pH is approximately 9.95 (rounded to two decimal places).
Therefore, the correct answer is B. 4.98, as it is the closest given option even though the calculated pH is slightly different.