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A solution of oxalic acid dihydrate (H:C2O4-2H:O) with a known concentration of 0.400 M H:C2O4-2H20 is titrated with a 0.333 M NaOH solution. How many L NaOH are required to reach the second equivalence point with a starting volume of 65.0 mL H:C2O4-2H2O , according to the following balanced chemical equation: H:C:O4 2H:0 + 2 NaOH – Na:C:O4 + 4 H:O mol H.GO. ZH:O 65.0 mL H:C2O4-2H2O 2 mol H:C2O4 2H:0 0.156 L NAOH LH:C:04-2H:0 1 molH.GO. ZH:O 0.333 mol NaOH -O.-2H:0 65.0 mL H:C:0.-2H:0 2 mol H:C:O.-2H:0 = 0.156 L NaOH 2H:0 1 mol H:C:O.-2H:0 0.333 mol NaOH ADD FACTOR DELETE ANSWER RESET *( ) 0.400 0.156 2 0.001 156 7.81 x 10 0.156 0.0781 0.333 4 39.0 78.1 1.56 x 10 1000 0.0520 5.20 x 10° 65.0 mol H:C204-2H:O g NaOH mL NaOH MH:C2O+-2H2O g H:C2O+-2H2O mol NaOH L NaOH M NaOH L H:C:O.-2H:O mL H:C2O4-2H:0

A solution of oxalic acid dihydrate (H:C2O4-2H:O) with a known concentration of 0.400 M H:C2O4-2H20 is titrated with a 0.333 M NaOH solution. How many L NaOH are required to reach the second equivalence point with a starting volume of 65.0 mL H:C2O4-2H2O , according to the following balanced chemical equation: H:C:O4 2H:0 + 2 NaOH – Na:C:O4 + 4 H:O mol H.GO. ZH:O 65.0 mL H:C2O4-2H2O 2 mol H:C2O4 2H:0 0.156 L NAOH LH:C:04-2H:0 1 molH.GO. ZH:O 0.333 mol NaOH -O.-2H:0 65.0 mL H:C:0.-2H:0 2 mol H:C:O.-2H:0 = 0.156 L NaOH 2H:0 1 mol H:C:O.-2H:0 0.333 mol NaOH ADD FACTOR DELETE ANSWER RESET *( ) 0.400 0.156 2 0.001 156 7.81 x 10 0.156 0.0781 0.333 4 39.0 78.1 1.56 x 10 1000 0.0520 5.20 x 10° 65.0 mol H:C204-2H:O g NaOH mL NaOH MH:C2O+-2H2O g H:C2O+-2H2O mol NaOH L NaOH M NaOH L H:C:O.-2H:O mL H:C2O4-2H:0

Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter14: Equilibria In Acid-base Solutions
Section: Chapter Questions
Problem 47QAP: A 0.4000 M solution of nitric acid is used to titrate 50.00 mL of 0.237 M barium hydroxide. (Assume...
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A solution of oxalic acid dihydrate (H2C2O4-2H2O) with a known concentration of 0.400 M H:C2O4 2H20 is titrated with a 0.333 M NAOH solution. How many L NaOH are required to reach the second equivalence point with a starting volume of 65.0 mL H2C2O4-2H2O , according to the following balanced chemical equation: H:C2O4 2H2O + 2 NaOH → Na:C204 + 4 H2O molH GO 2A:0 65.0 mL H2C2O4-2H2O 2 mol H2C2O4·2H2O 0.156 L NAOH LH:C204-2H2O molHGO 2A:O 0.333 mol NaOH C.Os 2H2O 65.0 mL H2C2O. 2H2O 2 mol H:C2O4·2H2O = 0.156 L NAOH -2H:O 1 mol H:C2O4 •2H:O 0.333 mol NaOH ADD FACTOR DELETE ANSWER RESET *( ) げ 0.400 0.156 2 0.001 156 7.81 x 10° 0.156 0.0781 0.333 4 39.0 78.1 1.56 x 104 1000 0.0520 5.20 x 10° 1 65.0 mol H2C2O4-2H2O g NaOH mL NaOH M H2C2O4 2H2O g H2C2O4-2H2O mol NaOH L NAOH M NaOH L H:C2O4-2H20 mL H2C2O4 2H2O
Transcribed Image Text:A solution of oxalic acid dihydrate (H2C2O4-2H2O) with a known concentration of 0.400 M H:C2O4 2H20 is titrated with a 0.333 M NAOH solution. How many L NaOH are required to reach the second equivalence point with a starting volume of 65.0 mL H2C2O4-2H2O , according to the following balanced chemical equation: H:C2O4 2H2O + 2 NaOH → Na:C204 + 4 H2O molH GO 2A:0 65.0 mL H2C2O4-2H2O 2 mol H2C2O4·2H2O 0.156 L NAOH LH:C204-2H2O molHGO 2A:O 0.333 mol NaOH C.Os 2H2O 65.0 mL H2C2O. 2H2O 2 mol H:C2O4·2H2O = 0.156 L NAOH -2H:O 1 mol H:C2O4 •2H:O 0.333 mol NaOH ADD FACTOR DELETE ANSWER RESET *( ) げ 0.400 0.156 2 0.001 156 7.81 x 10° 0.156 0.0781 0.333 4 39.0 78.1 1.56 x 104 1000 0.0520 5.20 x 10° 1 65.0 mol H2C2O4-2H2O g NaOH mL NaOH M H2C2O4 2H2O g H2C2O4-2H2O mol NaOH L NAOH M NaOH L H:C2O4-2H20 mL H2C2O4 2H2O
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