Question

A student used 0.230g of oxalic acid dissolved in 50.0 mL of water as a primary standard (MW 126.07 g/mol). 37.10 mL of sodium hydroxide aqueous solution were required to titrate the sample. What is the molarity of the sodium hydroxide solution? The neutralization reactions is:

Answer #1

m = 0.23 g of oxalic acid

MW = 126.07

V = 50 ml

then

V = 37.1 ml of NaOH si added

find molarity of NaOH that will neutralize it

then

H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(l)

we require a 2:1 ratio

then, 2 mol of base = 1 mol of acid

calculate mol of acid

mol = mass/MW = 0.23/126.07 = 0.00182438 mol of acid

since ratio is 2:1

we require

0.00182438*2 = 0.00364876 mol of base to neutralize it

for molarity

M = mol/L = 0.00364876/(37.1*10^-3) = 0.09834932614 M of NaOH

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