6th edition 12.25

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Jeremiah Hutauruk
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6th edition 12.25

Postby Jeremiah Hutauruk » Tue Jan 15, 2019 11:05 pm

12.25 Calculate the initial molarity of Ba(OH)2 and the molarities of Ba2 , OH , and H3O in an aqueous solution that contains 0.43 g of Ba(OH)2 in 0.100 L of solution.

Hai-Lin Yeh 1J
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Re: 6th edition 12.25

Postby Hai-Lin Yeh 1J » Tue Jan 15, 2019 11:17 pm

To calculate the initial molarity of Ba(OH)2, [remember molarity= mol/volume (L)], convert 0.43g into mol and divide by 0.100 because that's the volume:
0.43g (1mol/171.344g)(1/100) = 0.025 M Ba(OH)2
Then, if you know that it's in an aqueous solution, Ba(OH)2 will just dissociate and become ions: Ba(OH)2 -> Ba2+ + 2OH-
Thus, the molarity of Ba2+ ions should be the same since it's a one-to-one mole ratio, 0.025 M Ba2+. For the molarity of OH- ions, it's a 2:1 ratio, so you would multiply the molarity of Ba(OH)2 by 2: getting 0.050 M OH-

Lastly, to find the concentration of H3O+ ions, you know that [H3O+][OH-]=Kw=1.0 x 10^-14. Simply substitute the concentration of OH- to find the concentration of H3O+. You should get 2.0 x 10^-13 M

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