# Acid-Base Stoichiometry

1. Jul 4, 2009

### webz

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
n=MV

3. 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.

2. Jul 4, 2009

### Bohrok

You multiplied by 145 instead of 0.145 (or .145 as you had written) in your first line of work.

Last edited: Jul 4, 2009
3. Jul 4, 2009

### Staff: Mentor

No need for latex.

Simple math error.

4. Jul 4, 2009

### Staff: Mentor

Hard to say - s/he wrote .145 which can be short for 0.145.

5. Jul 4, 2009

### webz

It should have been 5.003 x10^-3, not just 5.003.

Answer is 46.32mL. Thanks for the help.