2. A 500.0 mL sample of H2SO4 solution was analyzed by taking a 100.0 mL aliquot

Question

2. A 500.0 mL sample of H2SO4 solution was analyzed by taking a 100.0 mL aliquot and adding 50.0 mL of 0.213 M NaoH. After the reaction occurred, an excess of OH ions remained in the solution. The excess base required 13.21 mL of 0.103 M HCl for neutralization. Calculate the concentration of the original sample of H2SO4 3. A student titrates 15.0 mL of an unknown acetic acid solution with 23.4 mL of a 0.423 M NaOH when he recognizes that he has gone well past the endpoint of the titration. What is the molarity of the acetic acid solution if it takes 7.80 mL of a 0.466 M HCI solution to neutralize the solution? 1 lt roguires 5 54 m

Explanation / Answer

The reaction between NaOH and HCl is

NaOH+ HCl -----àNaCl+ H2O

Moles of HCl = moles of NaOH= 0.103*13.21/1000 =0.001361

Volume of NaOH consumed= moles/ Molarity of NaOH= 0.001361/0.213=0.0064 L =0.0064*1000ml=6.4 ml

Volume of   NaOH used for titration with H2SO4= 50-6.4= 43.6 ml =43.6/1000L=0.0436L

Moles of NaOH in 0.0436L= 0.213*0.0436 =0.0093 moles

The reaction between H2SO4 and NaOH is

H2SO4+ 2NaOH ------à Na2SO4+ 2H2O

2 moles of NaOH requires 1 mole of H2SO4

0.0093 moles of NaOH requires 0.0093/2= 0.00465 moles of H2SO4

Hence molarity of H2SO4= Moles/ Volume in L = 0.00465/0.1= 0.0465M

2.

The reaction between HCl and NaOH is

NaOH+ HCl -----àNaCl+ H2O

1 mole of NaOH requires 1 mole of HCl for neutralization

Moles of HCl used = molarity* Volume in L =0.466*7.8/1000=0.003635

Moles of NaOH consumed =0.003635

Molarity of NaOH used = 0.423M, volume of NaOH used for neutralization with HCL= moles/Molairy

Volume of NaOH used = 0.003635/0.423=0.0086L= 0.0086*1000ml=8.6 ml

Volume of NaOH used for reaction with acetic acid = total volume- volume of NaOH used for titration with HCl = 23.4-8.6= 14.8 ml = 14.8/1000 L= 0.0148 L

Moles of NaOH used for titration with acetic acid = 0.423*0.0148 =0.0063

The reaction between acetic acid and NaOH is

CH3COOH+ NaOH ------àCH3COONa+ H2O

1 moles of acetic acid requires 1 mole of NaOH for neutralization

Hence moles of acetic acid = moles of NaOH= 0.0063

Molarity of acetic acid = moles of acetic acid/Volume in L= 0.0063*1000/15 =0.42 M

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