[scrum]

Hydrochloric acid (51.0 mL of 0.041 M) is added to 211.0 mL of 0.074 M Ba(OH)2 solution. What excess ions are left in the solution?
none
H+
OH - 


What is the concentration of the excess H+ or OH - ions left in this solution?
  M

First I got the mols of each thing
.051L * .041M = .002091Mol HCL

 .211L * .074M = .015614 Mol Ba(OH)2

Then i figured i'd need to mols of HCL to neutralize each mol of Ba(OH)2 as there is 1 H+ in HCL but 2 (OH)- in each Ba(OH)2.

So i multiplied the mols of HCL by 2 and got .004182.

since .015614 is greater than .004182, i figure that there are more OH- than H+, and this answer was correct. Now onto finding the concentration of left over OH-..

.015614 Mol (OH)2 - .004182 mol H+ = .011432 mol (OH)2 Left over

.011432 mol (OH2)/ .262(total liters)  = .043633 molarity (OH)- which is wrong...

can someone help me with what i did wrong?

[Sev]

Hydrochloric acid (51.0 mL of 0.041 M) is added to 211.0 mL of 0.074 M Ba(OH)2 solution. What excess ions are left in the solution?
none
H+
OH - 


What is the concentration of the excess H+ or OH - ions left in this solution?
  M

First I got the mols of each thing
.051L * .041M = .002091Mol HCL

 .211L * .074M = .015614 Mol Ba(OH)2

Then i figured i'd need to mols of HCL to neutralize each mol of Ba(OH)2 as there is 1 H+ in HCL but 2 (OH)- in each Ba(OH)2.

So i multiplied the mols of HCL by 2 and got .004182.

2 mol HCl reacts with 1 mol Ba(OH)₂, as you said.  So divide by 2 to get amount of Ba(OH)₂ that reacts.

[scrum]

i'm confused. divide my answer by 2?

[Sev]

So i multiplied the mols of HCL by 2 and got .004182.

You have to divide (not multiply) by 2 to get mol of Ba(OH)₂ that reacts.

[scrum]

I'm back to attempting this one again.

I did

.051L * 0.041 M to get .002091 mols HCL

then

.211 * .074 to get .015614 mols Ba(OH)2

Then i divided  .015614 by 2 and got .007807.

then i subtracted .002091 from .007807 and got .005716

then I took that over the total liters and got .02182

which is the same answer i had before that is incorrect. can someone tell me where I'm still going wrong?

[Borek]

Look at the net ionic reaction:

H⁺ + OH^- -> H₂O

You start with 2.091 milimole of H⁺ - that's OK

You start with 15.614 milimole of Ba(OH)₂ - how much OH^- is this?

[LQ43]

you should start by writing a balanced equation.

But its the net ionic equation that really matters

H+  +  OH- -->  H2O


your work:
"First I got the mols of each thing
.051L * .041M = .002091Mol HCL

 .211L * .074M = .015614 Mol Ba(OH)2

Then i figured i'd need to mols of HCL to neutralize each mol of Ba(OH)2 as there is 1 H+ in HCL but 2 (OH)- in each Ba(OH)2."

that was good


BUT   Ba(OH)2 --> Ba2+  +  2OH-
so the mols of OH- = 2 x mol Ba(OH)2 =  2 x .015614 = mol of OH- before

.051L * .041M = .002091Mol HCL = mol H+ before

use H+  +  OH- -->  H2O

to figure out how many of each reacted, subtract to see what is left after. Get conc of ions using total volume

set up a table to keep track of H+ and OH-,

   H+   OH-
before      
reacted      
after      

[LQ43]

sorry Borek, didn't see your reply before I posted...

[scrum]

so the mols of OH- = 2 x mol Ba(OH)2 =  2 x .015614 = mol of OH- before

OHHHHHHHHHHHHHHHHHHHHH!!!!!!

[Borek]

sorry Borek, didn't see your reply before I posted...

Happens