Calculate the hydronium ion concentration and the pH when

50.0mL of 0.40M NH3 is mixed with 50.0 mL of 0.40M HCl.

Showthe equation for the reaction in terms of the hydronium

ion.

4.98

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