If a 16.5 mL sample of 1.1 M solution of each of the following acids is reacted
ID: 1029667 • Letter: I
Question
If a 16.5 mL sample of 1.1 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)16.5 mL of HCl titrated with 0.85 M NaOH
volume of NaOH ________mL
total volume_________ mL
(b)16.5 mL of H2SO4 titrated with 0.85 M NaOH
volume of NaOH ________mL
total volume ________mL
(c)16.5 mL of HC2H3O2 titrated with 0.85 M NaOH
volume of NaOH _______mL
total volume ________ mL
Explanation / Answer
a. NaOH + HCl ---------------> NaCl + H2O
1 mole 1 mole
HCl NaOH
M1 = 1.1M M2 = 0.85M
V1 = 16.5ml V2 =
n1 =1 n2 = `1
M1V1/n1 = M2V2/n2
V2 = M1v1n2/M2n1
= 1.1*16.5*1/0.85*1 = 21.35 ml
The volume of NaOH = 21.35ml
total volume = 16.5 +21.35 = 37.85ml
b. H2So4 + 2NaOH -------------> Na2So4 + 2H2O
1 mole 2 moles
H2So4 NaOH
M1 = 1.1M M2 = 0.85M
V1 = 16.5ml V2 =
n1 = 1 n2 = 2
M1V1/n1 = M2V2/n2
V2 = M1V1n2/M2n1
= 1.1*16.5*2/0.85*1 = 42.7 ml
volume of NaOH = 42.7ml
total volume = 16.5 + 42.7 = 59.2ml
c.
HC2H3O2 + NaOH --------------> NaC2H3O2 + H2O
1 moles 1mole
HC2H3O2 NaOH
M1 = 1.1M M2 = 0.85M
V1 = 16.5ml V2 =
n1 = 1 n2 = 1
M1V1/n1 = M2V2/n2
V2 = M1V1n2/M2n1
= 1.1*16.5*1/0.85*1 = 21.35ml
The volume of NaOH = 21.35ml
total volume = 16.5 + 21.35 = 37.85ml