A solution is created by taking 42.34 mL of 0.3467-M sodium benzoate and 0.08712 L of 0.2222-M benzoic acid.
a. What is the pH of the solution if the Ka for benzoic acid is 6.281*10^-5?
b. What is the pH when 10.00 mL of 0.525 M NaOH is added to the above solution?
Acid base question?
2020-05-01 11:29 pm
回答 (1)
2020-05-02 12:27 am
a.
Initial moles of C₆H₅COOH = (0.3467 mol/L) × (42.34/1000 L) = 0.01468 mol
Initial moles of C₆H₅COO⁻ = (0.2222 mol/L) × (0.08712 L) = 0.01936 mol
Dissociation of benzoic acid:
C₆H₅COOH(aq) + H₂O(ℓ) ⇌ C₆H₅COO⁻(aq) + OH⁻(aq) Ka = 6.281 × 10⁻⁵
Henderson-Hasselbalch equation:
pH = pKa + log([C₆H₅COO⁻]/[C₆H₅COOH])
pH = -log(6.281 × 10⁻⁵) + log(0.01936/0.01468)
pH = 4.322
====
b.
Moles of OH⁻ added = (0.525 mol/L) × (10.00/1000 L) = 0.00525 mol
After addition of 0.00525 mol NaOH:
C₆H₅COOH(aq) + OH⁻(aq) → C₆H₅COOO⁻(aq) + H₂O(ℓ)
[C₆H₅COO⁻]/[C₆H₅COO]
= (Moles of HCOO⁻)/(Moles of HCOOH)
= (0.01936 + 0.00525)/( 0.01468 - 0.00525)
= 0.02461/0.00943
Henderson-Hasselbalch equation:
pH = pKa + log([C₆H₅COO⁻]/[C₆H₅COOH])
pH = -log(6.281 × 10⁻⁵) + log(0.02461/0.00943)
pH = 4.619
Initial moles of C₆H₅COOH = (0.3467 mol/L) × (42.34/1000 L) = 0.01468 mol
Initial moles of C₆H₅COO⁻ = (0.2222 mol/L) × (0.08712 L) = 0.01936 mol
Dissociation of benzoic acid:
C₆H₅COOH(aq) + H₂O(ℓ) ⇌ C₆H₅COO⁻(aq) + OH⁻(aq) Ka = 6.281 × 10⁻⁵
Henderson-Hasselbalch equation:
pH = pKa + log([C₆H₅COO⁻]/[C₆H₅COOH])
pH = -log(6.281 × 10⁻⁵) + log(0.01936/0.01468)
pH = 4.322
====
b.
Moles of OH⁻ added = (0.525 mol/L) × (10.00/1000 L) = 0.00525 mol
After addition of 0.00525 mol NaOH:
C₆H₅COOH(aq) + OH⁻(aq) → C₆H₅COOO⁻(aq) + H₂O(ℓ)
[C₆H₅COO⁻]/[C₆H₅COO]
= (Moles of HCOO⁻)/(Moles of HCOOH)
= (0.01936 + 0.00525)/( 0.01468 - 0.00525)
= 0.02461/0.00943
Henderson-Hasselbalch equation:
pH = pKa + log([C₆H₅COO⁻]/[C₆H₅COOH])
pH = -log(6.281 × 10⁻⁵) + log(0.02461/0.00943)
pH = 4.619
收錄日期: 2021-04-18 18:28:54
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