Prelab questions: From page 259.260: Q 1,2,3,6, 10,11. Note: Questions 10 and 11

Question

Prelab questions: From page 259.260: Q 1,2,3,6, 10,11. Note: Questions 10 and 11 needs more calculation. Answer on a separate sheet. All calculations should be shown with conversion factors, calculator value and final value with correct S.F. and unit.. Report sheet (Postlab) questions solve on this page 1. Calculate the molarity (M) of a solution that contains 4.58 g of H2C2O4. 2H2O (oxalic acid) in 350 mL. of solution. Show all steps with conversion factor and formula, equations etc. to get full credit ( Show molar mass calculation when used) 2 Calculate the volume (mL) of 0.350 M of KOH required to neutralize 25.51 mL of 0.155 M HCl solution. Show Equation and rearranged Equation before solving the problem. Show all steps with conversion factor and formula, equations etc.to get full credit.

Explanation / Answer

1)

m = 4.58 g H2C2O4*2H2O

V = 350 ml = 0.35 L

MW = 2(1) +2*12+4*16 + 2*(2*2 + 16) = 130 g/mol

mol = mass/MW = 4.58/130 = 0.0352 mol of H2C2O4*2H2O

1 mol of H2C2O4*2H2O --> 1 mol of H2C2O4

calculate molarity

M = mol/V = 0.0352 / 0.35 = 0.1

2)

V = ?

M = 0.35 KOH

V = 25.51 ml of M = 0.155 HCL

1 mol of acid + 1 mol of base = neutralization

find moles of acid used

M*V = mmol

mmol = (25.51*0.155 ) =3.954 mmol of acid

therefore, there are 3.954 mmol of base

mmol of base = M*V

V = mmol/M = 3.954/0.35 = 11.29 ml of base

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