Question

Titration (Acid-Base Reactions)

If a 18.0 mL sample of 1.5 M solution of each of the following acids is reacted with 0.85 M NaOH, how many milliliters of the NaOH are required for the titration? What is the total volume (in mL) of solution at the equivalence point?

(a) 18.0 mL of HCl titrated with 0.85 M NaOH

-volume of NaOH? mL

-total volume? mL

(b) 18.0 mL of HC₂H₃O₂ titrated with 0.85 M NaOH

-volume of NaOH? mL

-total volume? mL

(c) 18.0 mL of H₂SO₄ titrated with 0.85 M NaOH

-volume of NaOH? mL

-total volume? mL

Answer 1

The reaction between HCl and NaOH is given by

HCl+NaOH ----> NaCl + H2O

1mole of HCl reacts with 1mole of NaOH

The equivalence point for the reaction between HCl and NaOH can be calculated using the relation

M1V1=M2V2

M1= molarity of HCl, V1=Volume of HCl

M2= molarity of NaOH, V2 = Volume of NaOH

(a) 18.0 mL of HCl titrated with 0.85 M NaOH

M1V1=M2V2

M1= molarity of HCl = 1.5M, V1=Volume of HCl = 18.0 mL

M2= molarity of NaOH= 0.85M, V2 = Volume of NaOH=?

V2 = M1V1/M2 = 1.5*18/0.85 = 31.76 mL

Volume of NaOH = 31.76 mL

Total volume = volume of HCl+volume of NaOH=18+31.76 = 49.76mL

b) 18.0 mL of HC₂H₃O₂ titrated with 0.85 M NaOH

The reaction between HC2H3O2 and NaOH is given by

HC₂H₃O₂ + NaOH ------> C₂H₃O₂Na + H2O

1mole of HC₂H₃O₂ = 1mole of NaOH

M1V1=M2V2

M1= molarity of HC2H3O2 = 1.5M, V1=Volume of HC2H3O2 = 18.0 mL

M2= molarity of NaOH= 0.85M, V2 = Volume of NaOH=?

V2 = M1V1/M2 = 1.5*18/0.85 = 31.76 mL

Volume of NaOH = 31.76 mL

Total volume = volume of HC2H3O2+volume of NaOH=18+31.76 = 49.76mL

(c) 18.0 mL of H₂SO₄ titrated with 0.85 M NaOH

The reaction between H2SO4 and NaOH is given by

H2SO4 + 2NaOH -----> Na2SO4 + H2O

1mole of H2SO4 = 2moles of NaOH

The equivalence point can be calculated from the equation, M1V1/n1 = M1V2/n2

M1= molarity of H2SO4l = 1.5M, V1=Volume of H2SO4= 18.0 mL

M2= molarity of NaOH= 0.85M, V2 = Volume of NaOH=?

n1 = no. of moles of H2SO4 =1, n2= no. of moles of NaOH =2

V2 = M1V1*n2/M2*n1= 1.5*18*2/0.85*1 = 63.52 mL

Total volume of solution = 18+63.52 = 81.52 mL