Titration (Acid-Base Reactions) If a 18.0 mL sample of 1.5 M solution of each of
ID: 536962 • Letter: T
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
Explanation / Answer
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
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