Calculating Volume for Acid-Base Stoichiometry Problem

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of the volume of a sodium hydroxide (NaOH) solution needed to completely react with a given volume of phosphoric acid (H₃PO₄) solution. The focus is on stoichiometry and the proper application of molarity and volume in chemical reactions.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant presents a calculation involving the molarity of phosphoric acid and its volume to determine the moles of hydrogen ions produced.
  • Another participant identifies a multiplication error in the initial calculation, suggesting that the participant used 145 instead of 0.145 for the volume of phosphoric acid.
  • There is a mention of a potential misunderstanding regarding the notation used for the volume.
  • A later reply corrects the number of moles calculated, indicating it should be expressed as 5.003 x 10^-3 instead of 5.003.
  • One participant concludes that the final answer for the volume of NaOH required is 46.32 mL, thanking others for their help.

Areas of Agreement / Disagreement

Participants generally agree on the identification of a mathematical error in the initial calculation, but there is no consensus on the interpretation of the notation used for volume. The final answer is presented by one participant, but the discussion does not resolve the underlying confusion regarding the calculations.

Contextual Notes

The discussion highlights limitations in the initial calculations, particularly in the interpretation of volume units and the representation of numbers in scientific notation. There are unresolved aspects regarding the clarity of the notation used by the original poster.

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Homework Statement


Calculate the volume of 0.108 M NaOH solution required to react completely with 145 mL of phosphoric acid solution that is 1.15x10^-2 M.


Homework Equations


n=MV


The Attempt at a Solution


[tex]NaOH \rightarrow Na^{+}+OH^{-}[/tex]
[tex]H_{3}PO_{4} \rightarrow 3H^{+} + PO_{4}^{-}[/tex]

[tex]1.15x10^{-2} mol/L[/tex] [tex]H_{3}PO_{4} * .145L[/tex] [tex]H_{3}PO_{4} * 3 mol H^{+}/1 mol H_{3}PO_{4} = 5.003 mol H^{+}[/tex]

[tex]5.003 mol OH^{-} / .108 mol OH^{-} L = 46.324 L[/tex]

I clearly went wrong somewhere. Why in the world would I ever need 46 L to react with a 145 mL solution.

PS. I am bad with the latex reference codes, sorry.
 
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You multiplied by 145 instead of 0.145 (or .145 as you had written) in your first line of work.
 
Last edited:
No need for latex.

Simple math error.
 
Bohrok said:
You multiplied by 145 instead of 0.145 in your first line of work.

Hard to say - s/he wrote .145 which can be short for 0.145.
 
It should have been 5.003 x10^-3, not just 5.003.

Answer is 46.32mL. Thanks for the help.
 

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