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August 23, 2014

Homework Help: Chemistry II

Posted by Jayd on Wednesday, April 25, 2007 at 6:26pm.

Hydrochloric acid (77.0 mL of .267 M) is added to 236.0 mL of .0730 M Ba(OH)2 solution. What is the concentration of the excess H+ or OH- ions left in the solution?

Done two of these. They are all alike. Use the template from the previous problem. Post your work if you get stuck.

I can find the M of Ba(OH)2, but how do I get the concentration of the excess OH- ions?

Are you writing the equation before starting these problems? That is the secret to understanding what is going on.
Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O
mols Ba(OH)2 = 0.0730*0.236 = 0.0172 mols.
mols HCl = 0.077*0.267 = 0.0206

Now, which one is the "excess" reagent. We really need to do both of them; one to know which is in excess and the second one to make sure we are right.
Start with Ba(OH)2. We have 0.0172 mols. How many mols HCl will it take to completely neutralize it. It will be twice that or 0.0172 x 2 = 0.0344 and we DON'T HAVE THAT MUCH HCl. So we know HCl will be completely used up. But let's check it to make sure.

We have 0.0206 mols HCl. How much Ba(OH)2 will it take to completely neutralize it? It will take 0.0206/2 = 0.0103 mols Ba(OH)2. Do we have that much? Yes. And that agrees with the answer for the material we tried first.
So all of the HCl will be used and how much Ba(OH)2 will remain.
We had 0.0172 to start. We neutralize 0.0103. We have left 0.0172-0.0103 = ?? mols Ba(OH)2 remaining.
mols (OH^-) is twice that since there are two mols OH- in a mole of Ba(OH)2.
Then mols OH/liters (to obtain liters you want the total volume and that is 0.077 L + 0.2360 L = ?? LITERS).
Then determine pH.
Check my arithmetic. Check my work.

M = 4.41e-2
pH = 1.36

M = 4.41e-2
pH = 1.36

I was writing the equations but I was not completely understanding exactly what it was you were trying to explain. Thank you for your assistance.

Actually, the problem didn't ask for pH. Your value for (OH^-) is correct at 0.0441 M. BUT your value for pH actually is value for pOH.
pOH = -log(OH^-) = 1.36.
Then pH + pOH = 14 and
pH = 14-pOH = 14 - 1.36 = 12.64.

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