chem help?
2020-04-14 2:42 pm
What is the pH of the buffer solution in question 2 (Recall the values were: 2.0 L buffer of 0.24 M formic acid (HCOOH, Ka = 1.8x10-4) and 0.45 M sodium formate (NaHCOO)) after 0.050 moles NaOH is added?
回答 (1)
2020-04-14 2:52 pm
Initial moles of HCOOH = (0.24 mol/L) × (2.0 L) = 0.48 mol
Initial moles of HCOO⁻ = (0.45 mol/L) × (2.0 L) = 0.90 mol
After addition of 0.050 mol NaOH:
HCOOH(aq) + OH⁻(aq) → HCOO⁻(aq) + H₂O(ℓ)
[HCOO⁻]/[HCOOH]
= (Moles of HCOO⁻)/(Moles of HCOOH)
= (0.90 + 0.050)/(0.24 - 0.050)
= 0.95/0.19
Dissociation of HCOOH:
HCOOH(aq) + H₂O(ℓ) ⇌ HCOO⁻(aq) + H₃O⁺(aq) … Ka = 1.8 × 10⁻⁴
Henderson-Hasselbalch equation:
pH = pKa + log([HCOO⁻]/[HCOOH])
pH = -log(1.8 × 10⁻⁴) + log(0.95/0.19)
pH = 4.44
Initial moles of HCOO⁻ = (0.45 mol/L) × (2.0 L) = 0.90 mol
After addition of 0.050 mol NaOH:
HCOOH(aq) + OH⁻(aq) → HCOO⁻(aq) + H₂O(ℓ)
[HCOO⁻]/[HCOOH]
= (Moles of HCOO⁻)/(Moles of HCOOH)
= (0.90 + 0.050)/(0.24 - 0.050)
= 0.95/0.19
Dissociation of HCOOH:
HCOOH(aq) + H₂O(ℓ) ⇌ HCOO⁻(aq) + H₃O⁺(aq) … Ka = 1.8 × 10⁻⁴
Henderson-Hasselbalch equation:
pH = pKa + log([HCOO⁻]/[HCOOH])
pH = -log(1.8 × 10⁻⁴) + log(0.95/0.19)
pH = 4.44
收錄日期: 2021-04-18 18:27:32
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200414064208AA9bjAO