[maccha]

So, I'm learning about mixing strong acids and bases in unequal amounts and finding out which is in excess, and then finding the pH. For example, when mixing NaOH + HCl , H2O and NaCl results.. the OH and H+ ions have a one to one ratio so the calculation happens easily. But then in another practice question it mixes Ba(OH)2 and HCl, which when balanced I thought would look like this:
Ba(OH)2 + 2HCl --> 2H2O + BaCl2

So I assumed the ratio between H and OH would still be one to one, but when the text book does the calculations it says the ratio between OH and H is 2 to 1. Is there a different concept when mixing strong acids and bases? Help would be greatly appreciated!

[Astrokel]

It should be 1:1 if you are talking about OH^-:H⁺

[maccha]

Thanks for answering.. Okay so the actual question is : Calculate the pH resulting from mixing 25.0 mL of 0.0420 M Ba(OH)2 with 125.0 mL of 0.0120 M HCl.

The book then did dilution calculations like this:

0.0120 M HCL x 125.0ml/150.0 ml

But then for the dilution calculation for Ba(OH)2 it went

0.0420 M  Ba(OH)2 x 25.0 ml/150.0 ml x 2

It said it multiplied it by two because of the 2:1 mol ratio of OH- to H+, which leads back to my original question..

[Astrokel]

I'm confused, why is dilution involved?

[maccha]

Because we have to find the concentration of OH- and H+ to find pH, and since were mixing the two solution together they're going to be changed.

[macman104]

It should be 1:1 if you are talking about OH^-:H⁺

They react in a 1:1 ratio, but the reaction donates OH:H in a 2:1 ratio.  For Ba(OH)2 and HCl, the OH:H ratio is 2 to 1, because every one mole of Ba(OH)2 donates 2 OH anions.

If you write out those units, you will see that they are calculating the concentration of each ion in the final solution (assuming the volumes are additive).  First they find the moles and then divide by the new volume, so they multiply by two, because when you find the moles of Ba(OH)2, you will end up with 2 moles of OH anion.

[maccha]

Thanks so much! I fully understand now.

[Astrokel]

If you write out those units, you will see that they are calculating the concentration of each ion in the final solution (assuming the volumes are additive).  First they find the moles and then divide by the new volume, so they multiply by two, because when you find the moles of Ba(OH)2, you will end up with 2 moles of OH anion.

Why is OH:H ratio 2:1? Ba(OH)2 is dibasic while HCl is monoprotic but that is not the reason for multiplying 2 because the ratio of it should be 1:1. While i thought you multiplied it by 2 because like you said 1 mole of Ba(OH)2 contains 2 moles of OH- ion. Perhaps you could correct my thinking.

[Borek]

OH^- reacts with H⁺ in 1:1 ratio

Ba(OH)₂ reacts with HCl in 1:2 ratio

To convert [Ba(OH)₂] to [OH^-] you have to multiply the first by 2.

[Astrokel]

Borek, I agree with you but i don't understand what it mean't by OH:H is in a ratio of 2:1 because it is 1:1. Unless it is talking about 2 moles of OH^- per mole of Ba(OH)₂, but that is not linked to the ratio of OH^-:H⁺

[Borek]

i don't understand what it mean't by OH:H is in a ratio of 2:1 because it is 1:1

As you know what is OK - don't bother. There are so many erroneus ways of expressing correct ideas... it is not worth your time to straighten them all out

[macman104]

Borek, I agree with you but i don't understand what it mean't by OH:H is in a ratio of 2:1 because it is 1:1. Unless it is talking about 2 moles of OH^- per mole of Ba(OH)₂, but that is not linked to the ratio of OH^-:H⁺

When you react 1 mole of Ba(OH)2 with 1 mole of HCl, you get 2 moles of OH^- into solution and one mole of H⁺.