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0.4206 g of oxalic acid dihydrate (H_2 C_2 O_4 middot 2H_2 O) was weighed out and titrated with sodium hydroxide (NaOH) according to the procedure in the handout. V_i was recorded from the buret as 0.95 mL. V_f was recorded from the buret as 31.63 mL. (a) Calculate the volume of NaOH used from the buret. (b) Calculate the moles of oxalic acid dihydrate used. (Molar mass in manual, bottom p.41;use 4 sig figs!, i.e. 0.00XXXX moles) (c) Calculate the moles of sodium hydroxide needed to react with the oxalic acid used. See balanced equation in manual, p. 43. (don't forget to keep 4 sf !) (d) Calculate the molarity of the sodium hydroxide solution. (Remember molarity = moles/L). (remember 4 sig figs!)

Explanation / Answer

(a) volume of NaoH = Vf - Vi = (31.63 - 0.95 )ml = 30.68 ml

(b) moles = given weight / molecular weight

given weight = 0.4206 gm

molecular weight of oxalic acid dihydrated = (2*1 + 2*12 + 4*16 + 2*18 ) gm = 126 gm

moles of oxalic acid = 0.4206 gm / 126 gm = 0.003338

(c) standard reaction....

H2C2O4(aq) + 2NaOH(aq) --> Na2C2O4(aq) + 2H2O(l)

so it can be concluded from above equation that for titration of 1 mol oxalic acid requires 2 mol of NaOH.

hence, mole of NaOH will be = 2* 0.003338 = 0.006676

(d) molarity = moles / L

moles =  0.006676

volume = 30.68 ml = 0.03068 L

hence, molarity = 0.006676 / 0.03068 = 0.2176 M

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