0.140 grams of oxalic acid dihydrate (MW=126.068 g/mole) is dissolved in water in an Erlenmeyer flask. Remember that each mole of oxalic acid dissolved in water results in 2 moles of hydronium ion so it takes 2 moles of NaOH to neutralize one mole of oxalic acid. It requires 13.21 mLs of your prepared sodium hydroxide solution to reach the endpoint of the titration. What is the molarity of your NaOH solution?
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H2C2O4•2H2O dissolved in water
0.14g of the dihydrate = 1.11x10^-3moles
so, we have 1.12x10^-3moles H2C2O4
diprotic acid
H2C2O4 + 2NaOH --> NaC2O4 + 2H2O
1.11x10^-3moles H2C2O4 requires 2.22x10^-3moles OH-
2.22x10^-3 moles / 0.01321L = 0.168M
0.14g of the dihydrate = 1.11x10^-3moles
so, we have 1.12x10^-3moles H2C2O4
diprotic acid
H2C2O4 + 2NaOH --> NaC2O4 + 2H2O
1.11x10^-3moles H2C2O4 requires 2.22x10^-3moles OH-
2.22x10^-3 moles / 0.01321L = 0.168M