Calculating Volume for Acid-Base Stoichiometry Problem

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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:
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.