The following is the reaction showing the complete neutralization of calcium hydroxide (a base) with phosphoric acid: 3 Ca(OH)2 (aq) + 2 H3PO4 (aq) →>> Ca3(PO4)2 (s) + 6H₂O(l) A stock solution of calcium hydroxide (base) is made by dissolving 5.19 grams of it into some water and the volume is brought to 750 mL. A 50.0 mL portion of that stock solution is then titrated with a solution of 0.265 M phosphoric acid. How many milliliters (mL) of the acid are needed to completely neutralize the base in this reaction? (tolerance is ±0.1 mL) Type your answer...

Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter4: Chemical Reactions In Solution
Section: Chapter Questions
Problem 4.19QE: Write the net ionic equation for the reaction, if any, that occurs on mixing (a) solutions of sodium...
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The following is the reaction showing the complete neutralization of calcium hydroxide (a base) with phosphoric acid:
3 Ca(OH)2 (aq) + 2 H3PO4 (aq) →
Ca3(PO4)2 (s) + 6H₂O(l)
A stock solution of calcium hydroxide (base) is made by dissolving 5.19 grams of it into some water and the volume is brought to 750 mL. A 50.0 mL portion of that stock solution
is then titrated with a solution of 0.265 M phosphoric acid. How many milliliters (mL) of the acid are needed to completely neutralize the base in this reaction? (tolerance is ±0.1
mL)
Type your answer...
Transcribed Image Text:The following is the reaction showing the complete neutralization of calcium hydroxide (a base) with phosphoric acid: 3 Ca(OH)2 (aq) + 2 H3PO4 (aq) → Ca3(PO4)2 (s) + 6H₂O(l) A stock solution of calcium hydroxide (base) is made by dissolving 5.19 grams of it into some water and the volume is brought to 750 mL. A 50.0 mL portion of that stock solution is then titrated with a solution of 0.265 M phosphoric acid. How many milliliters (mL) of the acid are needed to completely neutralize the base in this reaction? (tolerance is ±0.1 mL) Type your answer...
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