[Ethereal]

30.0 mL of 0.08M HCl and 60 mL of 0.08M Ca(OH)₂ are combined. What is the pH of the final solution?

Please show work and explain!

[Fluorine]

Have you read the rules?

4. Please show that you've at least attempted the problem. We don't mind helping you solve problems but we are ethically opposed to doing homework for you. Violators will have their topic deleted or locked, and subject to banning.

1) Write out the balanced reaction.
2) Convert the HCl and Ca(OH)₂ into moles, respectively.
3) Is the acid a strong or weak acid?

3a) If weak, then I'd follow this method;
3a) If strong, then calculate which reagent is in excess and calculate the molarity of the hydroxide or hydrogen ion.

[Ethereal]

Sorry, I haven't read the rules yet; I'll do that now. 

As for the problem, it's not homework - it's from an exam I already took.  I got the question wrong and I've been trying to figure out the answer for the past hour!!!

1. I can't figure out the balanced equation.
2. 0.0024 moles of HCl and 0.0048 moles of Ca(OH)2, therefore, HCl is the limiting reagent.
3. Strong acid.

pH = -log[H₃O]

[Fluorine]

1) 2 HCl + Ca(OH)₂   CaCl₂ + 2 H₂O

How do you figure this out? Group 2 metal cations have 2+ charge. If you can't remember this then remember OH^- is 1- charge, so if calcium has two of them it's charge should be 2+. Chlorine, as well as other halogens will (almost) always have 1- charge, so you can figure out you'll need two for every one calcium. This is neutralization reaction (acid+base) so water formation is expected.

2) Correct, using the balanced equation you should be able to get pH from pOH.
3) Correct, it's strong because it fully dissociates so we don't need to consider Ka of HCl.

This is how I would approach it, though it was a bit ago I learned acid-base chemistry myself. Take it with a grain of salt.

[Ethereal]

That was the balanced equation I had on the test, but how do I derive the pH of the solution from that equation?  There aren't any OH or H3O molecules in the products, so how can I calculate their molarity?

And I've tried these equations but they (a) don't work or (b) are almost impossible (for me) to balance:

2H + 2Cl + Ca^2+ + OH + H20 ---> CaCl2 + H30 + H20  (This one is balanced but it doesn't help me)

and

HCl + Ca(OH)2 +H20 ---> CaCl2 + H30 + H20 (I can't balance this one to save my life!)

What am I doing wrong?

[Fluorine]

They are present - isn't Ca(OH)₂ in excess? Doesn't it dissociate OH^- completely (strong base)?

2. 0.0024 moles of HCl and 0.0048 moles of Ca(OH)2, therefore, HCl is the limiting reagent.

0.0024 mol HCl + 0.0048 mol Ca(OH)₂ = x CaCl₂ + H₂O and OH^- because you determined calcium hydroxide would in excess. Let's figure out by how much specifically; 0.0024mol HCl x 1 mol Ca(OH)₂/ 2mol HCl = 0.0012mol HCl

0.0012mol H⁺ and 0.0012mol  Cl^- will be present since two moles are required for both to form their respective products. Going off of this I would then subtract them 0.0048mol Ca(OH)₂ - HCl 0.0012mol = 0.0036mol Ca(OH)₂ unreacted/excess. Since it contains two OH^- per one calcium, that means it'll dissociate 0.0072mol OH^- for every 0.0036mol Ca²⁺ - basically 2:1 ratio.

Molarity is moles/volume (liters), so for OH^- we have 0.0072mol over 30mL + 60mL (see original question) which is 90mL or 0.09L. This gives us a concentration of 0.08M OH^-. Since we are using a strong base Kb is not necessary.

Let's find the pOH from molarity; pOH = -log₁₀[OH^-] so -log(0.08M) = 1.09691001. With two significant figures and rounded up, 1.1. Now that we have the pOH, do you recall any formulas that relate pOH and pH? How about...pH+pOH = 14 or re-written pH = 14-pOH? pH = 14-1.1 = 12.9. Let's do a quick logical check. A pH of 7< is considered basic. We know calcium hydroxide is a strong base and it was in excess, so 12.9 makes sense.


If I did it right, this should be correct. Always recheck my work to be safe.

[Ethereal]

Yes, that's the correct answer, and thank you very much for your assistance!

One question, though.  If the balanced equation is 2 HCl + Ca(OH)2   CaCl2 + 2 H2O, then where, exactly, is the hydroxide ion?  I see calcium chloride and water in the products, but where does the hydroxide ion fit in?  

[Ethereal]

Never mind!  I understand now!  Thanks a lot!!!

[Fluorine]

Glad to help.!

Just in case anyone else sees this and wonders the same thing, calcium hydroxide left over will dissociates the OH^-.

[Borek]

I guess it was enough to point out one of the substances is a limiting reagent, and excess is responsible for pH.