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#### kimi85

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##### calculating pH and
« on: September 01, 2007, 10:46:56 PM »
Calculate [H3O+] and pH in saturated Ba(OH)2 (aq), which contains 3.9 g Ba(OH)2 . 8 H2O per 100 ml solution.

What I did I first find the molarity then multiplied it to two. The I calculated the H+ using the ionization constant of water. But my answer is wrong. I calculated pH from pOH and my answer is wrong too. What is wrong with my computation?

The correct answer is 4 x 10-3 M H3O+ and pH is 12.4.

Thank you.

#### enahs

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##### Re: calculating pH and
« Reply #1 on: September 01, 2007, 11:21:26 PM »
Quote
The correct answer is 4 x 10-3 M H3O+ and pH is 12.4.

That is impossible.

What would be a pH of a solution with 4x10-3 M H3O+?

#### Borek

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##### Re: calculating pH and
« Reply #2 on: September 02, 2007, 04:21:48 AM »
Show your numbers, approach (through pOH) seems correct. The only obvious mistake I can think off is molar mass of hydrated hydroxide - have you forgot water, or not?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### kimi85

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##### Re: calculating pH and
« Reply #3 on: September 02, 2007, 09:16:10 PM »
I'm sorry for the error. It should be 4 x 10-13 M H3O+

I also included the molecular weight of the hydrate but I'm still wrong

Here's what I did:

(3.9/315.46)/0.1 L = 0.1236 x 2 = 0.24725 pOH = 0.60686 pH = 14-pOH = 13.39

3O+:

1 x 10-14/0.24724 = 4 x 10-14

#### enahs

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##### Re: calculating pH and
« Reply #4 on: September 02, 2007, 09:37:57 PM »
The answer you are giving as the correct answer for the conditions you are giving are for 0.39 g of Ba(OH)2.

However 3.6 g is much closer to the real solubility value.

It looks like when working the solution they forgot to divide by 0.1L.

#### kimi85

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##### Re: calculating pH and
« Reply #5 on: September 02, 2007, 10:05:12 PM »
okay. thank you.