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Tasya [4]
3 years ago
9

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 th

e solution upon neutralization of half of the weak base. The Kb of the base is 1.5×10-6.
Chemistry
1 answer:
Pavel [41]3 years ago
4 0

Answer:

pH = 8.18

Explanation:

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

X + HCl → HX⁺ + Cl⁻

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

Now, using H-H equation for bases:

pOH = pKb + log [XH⁺] / [X]

<em>Where pOH is the pOH of the buffer (pH = 14 -pOH)</em>

<em>pKb is -log Kb = 5.824</em>

<em>And [X] [HX⁺] are the molar concentrations of each specie</em>

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

<h3>pH = 8.18</h3>
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