Consider the following two reactions involving oxalic acid.
2MnO4− + 5H2C2O4 + 6H+ → 2Mn2+ + 10CO2 + 8H2O |
H2C2O4 + 2OH− → C2O42− + 2H2O |
What volume of 0.0100 M KMnO4 solution will titrate a solution prepared from 46.0 mg of H2C2O4?
What volume of 0.0100 M NaOH solution will titrate a solution
prepared from 46.0 mg of
H2C2O4?
1)
Balanced chemical equation is:
2MnO4− + 5H2C2O4 + 6H+ → 2Mn2+ + 10CO2 + 8H2O
Molar mass of H2C2O4,
MM = 2*MM(H) + 2*MM(C) + 4*MM(O)
= 2*1.008 + 2*12.01 + 4*16.0
= 90.036 g/mol
mass(H2C2O4)= 0.046 g
use:
number of mol of H2C2O4,
n = mass of H2C2O4/molar mass of H2C2O4
=(4.6*10^-2 g)/(90.04 g/mol)
= 5.109*10^-4 mol
According to balanced equation
mol of MnO4- reacted = (2/5)* moles of H2C2O4
= (2/5)*5.109*10^-4
= 2.044*10^-4 mol
This is number of moles of MnO4-
use:
M = number of mol / volume in L
0.01 = 2.044*10^-4/ volume in L
volume = 0.0204 L
volume = 20.4 mL
Answer: 20.4 mL
2)
Balanced chemical equation is:
H2C2O4 + 2OH− → C2O42− + 2H2O
Molar mass of H2C2O4,
MM = 2*MM(H) + 2*MM(C) + 4*MM(O)
= 2*1.008 + 2*12.01 + 4*16.0
= 90.036 g/mol
mass(H2C2O4)= 0.046 g
use:
number of mol of H2C2O4,
n = mass of H2C2O4/molar mass of H2C2O4
=(4.6*10^-2 g)/(90.04 g/mol)
= 5.109*10^-4 mol
According to balanced equation
mol of OH- reacted = (2/1)* moles of H2C2O4
= (2/1)*5.109*10^-4
= 1.022*10^-3 mol
This is number of moles of OH-
use:
M = number of mol / volume in L
0.01 = 1.022*10^-3/ volume in L
volume = 0.102 L
volume = 102 mL
Answer: 102 mL
Consider the following two reactions involving oxalic acid. 2MnO4− + 5H2C2O4 + 6H+ → 2Mn2+ +...
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