Titration of Acetic Acid with Sodium Hydroxide In these reactions sodium hydroxi

Question

Titration of Acetic Acid with Sodium Hydroxide In these reactions sodium hydroxide is the base and reacts with acetic acid in the vinegar We can consider this to be a general reaction for a generic acid with one acidic proton called HA or be more specific and use the acetic acid formula (HC)2 H_3 O_2) as shown. General: HA + NaOH rightarrow H_2 O + NaA Specific: HC_2 H_3 O_2 + NaOH rightarrow H_2 O + NaC_2 H_3 O_2 What is important from these equations is that the sodium hydroxide and the acetic acid on a 1:1 mole basis (one mole of NaOH reacts with one mole of acid).

Explanation / Answer

Trial 1:

Concentration of acid:

mmol of base = MV = 34.8*0.25= 8.7 mmol of NaOH

then, due to 1:1 stoichiometry

mmol of acid = mmol of NaOH = 8.7

M = mol/V

M = mmol/mL = 8.7 / 10 = 0.87 M

apply same logic for each trial:

Trial 2:

M = mmol/mL = (M1*V1)/Vaci d= 0.25*V1/10 = 0.025 * V1 = 0.025*34.56 = 0.864 M of acid

Trial 3:

M = mmol/mL = (M1*V1)/Vaci d= 0.25*V1/10 = 0.025 * V1 = 0.025*35.32 = 0.883 M of acid

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.