What volume of 0.700 M KOH is required to react completely with 0.605 L of 0.850 M H3PO4

i got .735L but i know its wrong. help please? what did i do wrong?

"I don't know what you did wrong, but i know how to do it right."

Heh.

Here's the correct solution:

(0.850 mol/L) (0.605 L) = 0.51425 mol of H3PO4

3KOH + H3PO4 ---> K3PO4 + 3H2O

For every one H3PO4 reacted, you need three KOH

0.51425 mol times 3 = 1.54275 mole of KOH

1.54275 mol divided by 0.700 mol/L = 2.20 L (rounded to three sig figs)

More:

http://www.chemteam.info/AcidBase/Titrat…

Here's an interesting question asked by Chase Haynes:

http://answers.yahoo.com/question/index;…

with an interesting manner of how the answer was presented.

"and what would happen if H3PO4 has 3 acidic protons?"

That's why the 3KOH are required for one H3PO4. Each of the three acidic hydrogens in H3PO4 is reacted with one of the three hydroxides provided by the three KOH.

I don't know what you did wrong, but i know how to do it right. Thats more important. There actually multiple ways to solve this. But its easiest setting it up with cross multiplication, as in a proportion.

0.700M/x = 0.850M/0.605L Solve for x

x= (0.700Mx0.605L) / 0.850M Molarity cancels out and you are left with liters of KOH