Calculating Volume for Acid-Base Stoichiometry Problem

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To calculate the volume of 0.108 M NaOH needed to react with 145 mL of 1.15x10^-2 M phosphoric acid, the stoichiometry of the reaction must be considered. The initial calculation incorrectly used 145 instead of 0.145 for the volume of the acid, leading to an erroneous result of 46 L. The correct number of moles of H+ produced from phosphoric acid is 5.003 x 10^-3, which should then be used to find the volume of NaOH required. After correcting the calculations, the required volume of NaOH is determined to be 46.32 mL. This highlights the importance of careful unit conversion in stoichiometry problems.
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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


NaOH \rightarrow Na^{+}+OH^{-}
H_{3}PO_{4} \rightarrow 3H^{+} + PO_{4}^{-}

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

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

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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