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 CaCl
2 + H
2O
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 Ca
2+ - 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 = -log
10[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.