Question

Question 7 2 pts Consider the following sodium hydroxide, oxalic acid titration data. Remember sodium hydroxide is in the buret mass of oxalic acid - 1.301 g initial buret reading = 2.16 mL final buret reading = 39.13 mL Calculate the molarity of the sodium hydroxide. .761 M 0.705 M 733 M 782M

Answer 1

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COV (09) Oxalic acid (904) ,(NaOH) mbodium hydroxide The balance equation, 2 NaOH + H₂C₂O4 can, Na, C04 cog) + 2 H₂O (1) Oralie alid & Mass = 1-3019 & molar mass = 90.03 g/mol no. of moles = mass (9 - = 1-3019 molar mass (g/mol) 90.03 g/mol - from equation, 2 mol Noot rearts with Imol CH₂O2 so - mol NaOH rearts with 0.0144 mol Cette of 2 mol x 0.0144 mol = 0.0289 mot of neadh. I mol Now, calculate the molarity Nook Molarity - no. of moles of solule (mol) volume of solution (4) Volume = final reading - initial reading - 39013 mL-2.16mL. = 36.97mL = 0.036974 = 0.782 mol/l = 0.782 M 0.03697 L option (6 0.762 M is Correct answer [NooH] = 0.0289 moh.