Question

A 31.25 mL aliquot of weak base that has a concentration of 0.683 M will be titrated with 0.434 M HCl. Calculate the pH of of the solution upon neutralization of half of the weak base. The Kb of the base is 1.5×10-6.

Answer 1

pH = 8.18

Step-by-step explanation:

The weak base, X, reacts with HCl as follows:

X + HCl → HX⁺ + Cl⁻

Where 1 mole of X with 1 mole of HCl produce 1 mole of HX⁺ (The conjugate acid of the weak base).

Now, using H-H equation for bases:

pOH = pKb + log [XH⁺] / [X]

Where pOH is the pOH of the buffer (pH = 14 -pOH)

pKb is -log Kb = 5.824

And [X] [HX⁺] are the molar concentrations of each specie

Now, at the neutralization of the half of HX⁺, the other half is as X, that means:

[X] = [HX⁺]

And:

pOH = pKb + log [HX⁺] / [X]

pOH = 5.824 + log 1

pOH = 5.824

pH = 14-pOH

pH = 8.18