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0.250 molH PO LH3PO4 0.250 mol H₂PO4 LH₂PO₂ X 3 mol NaOH 1 mol H₂PO4 3 1 A solution of phosphoric acid (H3PO4) with a known concentration of 0.250 M H3PO4 is titrated with a 0.800 M NaOH solution. How many mL of NaOH are required to reach the third equivalence point with a starting volume of 72.0 mL H3PO4 according to the following balanced chemical equation: mol NaOH molH3PO4 72.0 mL H₂PO4 0.800 mol NaOH = 67.5 mL NaOH 0.800 43.2 X 0.250 H3PO4 + 3 NaOH → Na3PO4 + 3 H₂O 72.0 0.800 0.0675 1000 MH3PO4 mol H3PO4 mL H3PO4 M NaOH mL NaOH mol NaOH 67.5 6.75 x 104 g NaOH g H3PO4 1 22.5 = 0.001 67.5 72.0 LH3PO4 mol NaOH mL NaOH 3 7.50 mL H3PO4 LNaOH 2

0.250 molH PO LH3PO4 0.250 mol H₂PO4 LH₂PO₂ X 3 mol NaOH 1 mol H₂PO4 3 1 A solution of phosphoric acid (H3PO4) with a known concentration of 0.250 M H3PO4 is titrated with a 0.800 M NaOH solution. How many mL of NaOH are required to reach the third equivalence point with a starting volume of 72.0 mL H3PO4 according to the following balanced chemical equation: mol NaOH molH3PO4 72.0 mL H₂PO4 0.800 mol NaOH = 67.5 mL NaOH 0.800 43.2 X 0.250 H3PO4 + 3 NaOH → Na3PO4 + 3 H₂O 72.0 0.800 0.0675 1000 MH3PO4 mol H3PO4 mL H3PO4 M NaOH mL NaOH mol NaOH 67.5 6.75 x 104 g NaOH g H3PO4 1 22.5 = 0.001 67.5 72.0 LH3PO4 mol NaOH mL NaOH 3 7.50 mL H3PO4 LNaOH 2

Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter15: Additional Aqueous Equilibria
Section: Chapter Questions
Problem 93QRT: When 40.00 mL of a weak monoprotic acid solution is titrated with 0.100-M NaOH, the equivalence...
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0.250 molH3PO4 LH3PO4 0.250 mol H₂PO4 LH₂PO4 X 3 mol NaOH 1 mol H₂PO4 3 1 A solution of phosphoric acid (H3PO4) with a known concentration of 0.250 M H3PO4 is titrated with a 0.800 M NaOH solution. How many mL of NaOH are required to reach the third equivalence point with a starting volume of 72.0 mL H3PO4 according to the following balanced chemical equation: mol NaOH molH3PO4 72.0 mL H3PO4 0.800 mol NaOH = 67.5 mL NaOH 0.800 43.2 X 0.250 H3PO4 + 3 NaOH Na3PO4 + mL H3PO4 72.0 0.800 0.0675 1000 M H3PO4 mol H3PO4 67.5 mol NaOH 6.75 x 104 M NaOH mL NaOH g NaOH g H3PO4 1 22.5 = 3 H₂O 0.001 67.5 72.0 LH3PO4 mol NaOH mL NaOH 3 7.50 mL H3PO4 L NaOH 2
Transcribed Image Text:0.250 molH3PO4 LH3PO4 0.250 mol H₂PO4 LH₂PO4 X 3 mol NaOH 1 mol H₂PO4 3 1 A solution of phosphoric acid (H3PO4) with a known concentration of 0.250 M H3PO4 is titrated with a 0.800 M NaOH solution. How many mL of NaOH are required to reach the third equivalence point with a starting volume of 72.0 mL H3PO4 according to the following balanced chemical equation: mol NaOH molH3PO4 72.0 mL H3PO4 0.800 mol NaOH = 67.5 mL NaOH 0.800 43.2 X 0.250 H3PO4 + 3 NaOH Na3PO4 + mL H3PO4 72.0 0.800 0.0675 1000 M H3PO4 mol H3PO4 67.5 mol NaOH 6.75 x 104 M NaOH mL NaOH g NaOH g H3PO4 1 22.5 = 3 H₂O 0.001 67.5 72.0 LH3PO4 mol NaOH mL NaOH 3 7.50 mL H3PO4 L NaOH 2
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