Answer:
4.98
Step-by-step explanation:
In the presence of the strong acid HCl, the weak base NH₃ will react completely producing the salt NH₄Cl.
The ion NH₄⁺ being the conjugate acid of a weak base will affect the pH after this neutralization by hydrolyzing water:
NH₄⁺ + H₂O ⇆ H₃O⁺ + NH₃
Thus set up the ICE table to solve this question in the customary manner, but first compute the concentration of NH₄⁺ produced in the neutralization reaction:
NH₃ + HCl ⇒ NH₄⁺ + Cl⁻
V = 50.0 ml x l L/ 1000 mL = 0.05 L ( Need V in L for units consistency )
mol NH₃ to react : 0.05 L x 0.40 mol/L = 2.00 x 10⁻²
mol HCl to react : 0.05 L x 0.40 mol/L = 2.00 x 10⁻²
Since 1 mol HCl react with 1 mol NH₃ to produce 1 mol NH₄⁺, the amount of NH₄⁺ produced will be 2.00 x 10⁻² mol and its concentration is:
V sol = 0.05 L + 0.05 L = 0.10 L
Molarity NH₄⁺ = mol/ V sol = 2.00 x 10⁻² mol/ 0.100 L = 0.2 M
Now we are ready to set up ICE table:
NH₄⁺ + H₂O ⇆ H₃O⁺ + NH₃
I 0.2 0 0
C -x +x +x
E 0.2-x x x
Kₐ = [H₃O⁺][NH₃]/[NH₄⁺] = x² / (0.2 -x )
Since we know NH₄⁺ is a weak acid, we can make the approximation 0.2 - x ≈ 0.2
Ka for NH₄⁺ = Kw/ Kb NH₃
Kb NH₃ = 1.8 x 10⁻⁵ ⇒ Ka NH₄⁺ = 10⁻¹⁴ / 1.8 x 10⁻⁵ = 5.56 x 10⁻¹⁰
Now we solve for x:
5.56 x 10⁻¹⁰ = x²/0.2 ⇒ x = √(5.56 x 10⁻¹⁰ x 0.2) = √1.11 x 10⁻¹⁰
x = 1.05 x 10⁻⁵
( Indeed x is only 1.05 x 10⁻⁵ is neglible compared to 0.2. This approximation is good up to 5 %, if greater we will have to solve the quadratic equation)
Now that we know [H₃O⁺] = 1.05 x 10⁻⁵, pH is given by
pH = - log [H₃O⁺] = - log 1.05 x 10⁻⁵ = 4.98