12.25 6th Edition

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Joussie Camacho 4I
Posts: 33
Joined: Tue Oct 02, 2018 12:16 am

12.25 6th Edition

Postby Joussie Camacho 4I » Sat Dec 08, 2018 11:40 am

It asks to calculate the initial molarity of Ba(OH)2 and the molarities of Ba+, OH-, and H3O+ in an aqueous solution that contains 0.43 g Ba(OH)2 on 0.100 L of solution. I don't understand how to find the molarity of H3O+ because I don't know the mole ratio since I thought that the equation would just be Ba(OH)2 +H2O --> Ba^2+ + 2OH-

Nghi Nguyen 2L
Posts: 61
Joined: Fri Sep 28, 2018 12:18 am

Re: 12.25 6th Edition

Postby Nghi Nguyen 2L » Sat Dec 08, 2018 10:01 pm

You would be able to find the molarity of H3O+ by getting molarity of OH- and then converting it using [H3O+][OH-] = 1.00 x 10^-14.

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