Explanation on Chemistry?

[匿名]

2021-03-29 9:54 am

0.4 mol of NaOH(s) is added to 1.00 L of a 1.00 M solution of an acid with KA = 1·10-4. Calculate the pH of the resulting solution in two sig. figs. Assume the volume of the solution does not change when the NaOH(s) is added.

回答 (1)

Uncle Michael

2021-03-29 10:05 am

✔ 最佳答案

Denote the acid as HA.

The addition of each mole of NaOH converts 1 mole of HA to form 1 mole of A⁻ ions.
Moles of HA after addition = (1.00 mol/L) × (1.00 L) - (0.4 mol) = 0.6 mol
Moles of A⁻ ions after addition = 0.4 mol

After addition of NaOH:
[A⁻]/[HA] = (Moles of A⁻ ions) / (Moles of HA) = 0.4/0.6

Consider the dissociation of HA:
HA(aq) + H₂O(ℓ) ⇌ H₃O⁺(aq) + A⁻(aq)

Henderson-Hasselbalch equation:
pH = pKa + log([A⁻]/[HA])
pH = -log(1 × 10⁻⁴) + log(0.4/0.6)
pH = 3.8

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