Calculate the pH of 100mL HOAc + 100mL Ba(OH)2

[Joules23]

Homework Statement

find the pH of:
100mL x .20M HOAc + 100mL x.10M Ba(OH)2

The Attempt at a Solution

100x.2 = 20mmol HOAc
100x.1 = 10mmol Ba(OH)2
20-10=10
10/200=.05M
pOH = -log(.05) = 1.3
pH=14-1.3=12.7

Is this correct?

 

[Borek]

No. Ba(OH)2 is 'diprotic'.

 

[chemisttree]

Ba(OH)2 is not diprotic. Dibasic.

Think in terms of equivalents instead of moles (or mmol) with the barium hydroxide...

BTW, the way you did the work, it wasn't clear (to you) whether the "10" left over in the reaction was barium or acetic acid. Using the logic you provided, the acetic acid was actually present in excess and the pH should have been less than 7.

Using equivalents in place of moles should "completely neutralize" any misunderstanding you may have.

 

[GCT]

Homework Statement

find the pH of:
100mL x .20M HOAc + 100mL x.10M Ba(OH)2

The Attempt at a Solution

100x.2 = 20mmol HOAc
100x.1 = 10mmol Ba(OH)2
20-10=10
10/200=.05M
pOH = -log(.05) = 1.3
pH=14-1.3=12.7

Is this correct?

hint, hint, the situation is going to be "neutralized" as chemisttree has said

However, the conjugate of the acetic acid is a base with a significant Kb, you're going to need to use the Kb equation to find the [OH-], then the pH through further calculations.

 

[Borek]

Ba(OH)2 is not diprotic. Dibasic.

That's what happens when English is not your first language... :grumpy: