I don't get this Titration part. Is there any formula to solve this?

Question

Hello, I'm trying to figure this one out:
"Calculate the volume (mL) of 0.100 M H3PO4 required to neutralize 30.0 mL of 0.050 M Ca(OH)2."

Am I supposed to use a certain formula for this? How do I do this problem? Where should I start this? I read the text book but it doesn't clearly explain How I do this.

Answer 1

Start with the balanced equation for the reaction. This one is kinda tricky with the polyprotic (phosphoric) acid and calcium but the products are going to be water and calcium hydrogen phosphate.

H3(PO4) + (Ca(OH)2) => 2 H2O + CaH(PO4)

Now you start with 30.0 ml of Ca(OH)2, or 0.0300 L.

And using the concentration of that solution you can compute the # of moles of Ca(OH)2 in that amount of solution.

(0.0300 L) x (0.050 M Ca(OH)2/L = 0.015 Moles of Ca(OH)2

Now the ratios from the balanced equation. One mole of Ca(OH)2 reacts with one mole of H3(PO4). So an equal amount of one is required to neutralize the other.

0.015 Moles of H3(PO4)

Finally the volume. If you have 0.015 Moles of H3(PO4) and its in a 0.100 Moles of H3(PO4)/L solution, you just divide to get the volume.

[ 0.015 Moles of H3(PO4) ] / [ 0.100 Moles of H3(PO4)/L ] = 0.015 L of H3(PO4) solution, or 15 ml of H3(PO4) solution.


Having computed the answer you can see, that there is a simple ratio you can use to get the answer as well. From balancing the equation you know that there is a 1:1 ratio for the two compounds in the question. The Ca(OH)2 has half the concentration of the H3(PO)4.

So you could reason that it would take half as much of the H3(PO)4 with twice the concentration to neutralize, or 30 ml / 2 = 15 ml.

Don’t forget your significant figures. Several numbers were reported to 3 sig figs, but the concentration of Ca(OH)2 was reported to 2, so your answer should have 2 sign figs.