Question

A 38.22 mL aliquot of weak acid that has a concentration of 0.882 M will be titrated with 0.289 M NaOH. Calculate the pH of of the solution upon neutralization of half of the weak acid. The Ka of the acid is 6.8×10-7.

Answer 1

pH = 6.167

Step-by-step explanation:

The weak acid, HX, reacts with NaOH as follows:

HX + NaOH → NaX + H₂O

Where 1 mole of HX with 1 mole of NaOH produce 1 mole of NaX (The conjugate base of the weak acid).

Now, using H-H equation:

pH = pKa + log [NaX] / [HX]

Where pH is the pH of the buffer

pKa is -log Ka = 6.167

And [NaX] [HX] are the molar concentrations of each specie

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

[NaX] = [HX]

And:

pH = pKa + log [NaX] / [HX]

pH = 6.167 + log 1

pH = 6.167