CHEM 1411 - E nt 9 Determination of HaSO Roll # Name: Section: Date: Report Shee

Question

CHEM 1411 - E nt 9 Determination of HaSO Roll # Name: Section: Date: Report Sheet (Data must be recorded in pen) Molarity of the NaOH solution used -0.106 n- Titration of NaOH vs H2SO4 Trial 3 neces Trial 2 (Pilot) Trial 1 _pfw 1. Volume of sulfuric acid 1000ml )an, solution pipetted 2. Initial buret reading 3. Final buret reading 4. Volume of NaOH 5. Moles of NaOH 6. Moles H2SO4 neutralized by NaOH* 7. Molarity of H2SO4 in sulfuric acid solution* 8. Mass of H2SO4 in 100.0 mL of sulfuric acid solution* Equation for the neutralization reaction (*Show calculations for steps 5-8)

Explanation / Answer

Q6

1 mol of NaOH neutralizes = 1/2 mol of H2SO4 since it is dirptoci

mmol of NaOH = MV = 0.1*(3-0) = 0.3 mmol of NaOH

then, 0.3/2 = 0.15 mmol of H2SO4 present

Q7.

Molarity of acid is given by

[H2SO4] = mmol /V= 0.15 / 10 = 0.015 M

Q8

mass of acid in V = 100 mL

mol = MV = 0.015*0.1 = 0.0015 mol of acid

mass = mol*MW = 0.0015*98 = 0.147 g of acid in 100 mL

equation is given by

H2SO4 + 2NaOH = 2H2O + Na2SO4

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---