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Molarity of OH- ions when Ba(OH)2 is added to NaOH

Submitted by hey.twix on Wed, 06/24/2009 - 12:20

I have another doubt regarding concenration terms...

What volume of 0.40M Ba(OH)2 must be added to 50.0 mL of 0.30M NaOH to get a solution in which the molarity of the OH- ions is 0.50M?

Thank you!

I still do not get how to do this problem...
any hints anyone?

I looked at this last night and didn't get very far. Perhaps my thought may help you.
Let the volume of Ba(OH)2 be y mL
Total volume of solution (Ba(OH)2 and NaOH) = (50 + y) mL
Moles NaOH added = 0.015 so this contains 0.015 moles OH- ions. This is obtained by multiplying molarity by volume in Litres.
Moles Ba(OH)2 = (y/1000) x 0.40 = 0.0004y moles
Moles OH- in Ba(OH)2 = 0.0008y moles
Final solution has a molarity of 0.50 moles/litre for OH- ions
So 1000 mL contains 0.50 moles OH- ions
So (50 + y) mL contain (0.50/1000) x (50 + y) = 0.0005(50 + y) moles = 0.025 + 0.005y moles
Also this must equal 0.015 + 0.0008y moles of OH- ions

0.025 + 0.005y = 0.015 + 0.0008y
0.005y - 0.0008y = 0.015 - 0.025

This is where I come unstuck as I will have a negative value. Perhaps you can see what the solution is going to be.

I have played around and found the answer of 12.5 mL just by putting values in and calculating. However, I can't see the method.
12.5 mL Ba(OH)2 contains (12.5/1000) x 0.4 moles Ba(OH)2 = 0.005 moles Ba(OH)2
This is equivalent to 2 x 0.005 = 0.010 moles OH- ions
If this is added to 50 mL 0.3 mole NaOH which contains 0.015 moles OH- ions, the total number of moles of OH- ions will be 0.025 moles OH- in a total volume of 50 + 12.5 mL = 62.5 mL
This gives an OH- concentration of 0.025/0.0625 = 0.4 moles/ Litre

I hate to be beaten but still am. However, I tried the following which almosts gives the right answer.
Total moles OH- = (0.015 + 0.0008y) moles
Total volume = (50 + y) mL
So molarity, 0.4 = (0.015 + 0.0008y)/(50 + y)/1000
0.4(50 + y)/1000 = (0.015 + 0.0008y)
0.4(50 + y) = 1000(0.015 + 0.0008y)
20 + 0.4y = 15 + 0.8y
0.4y = 5
y = 1.25 mL (not the 12.5 expected). Have a look at this solution and see if you can find an error. It is 22.50 here and my brain is struggling. Hope this helps. I'll watch out to see how you solve it.

V1*C1 + V2*C2 =  0.50
      V1 + V2           

V1 is the volume that you are looking for
C1 is the concentration of OH- in first solution = 0.80M  (0.40M * 2)
V2 is the volume of the 2nd solution = 50 ml
C2 is the concentration of 2nd solution = 0.30M

solve for V1 and you get 33 1/3 ml

you do not need to change volumes to liters since all volumes are in ml your answer will be in ml.

Thanks Valdo
I spent a long time on this one.

Thank you very much! That was such a simple solution...