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Topic: Help with Ksp  (Read 4946 times)

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emerson

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Help with Ksp
« on: May 22, 2006, 12:05:11 AM »
I've already found the Ksp of BaCO3, but now I'm supposed to "use my experimental value of Ksp to determine the solubility of BaCO3" I got he equation 0=x2+0.1004-5.9 x 10-5.  Now my question is, am I supposed to use quadratic formula or is there another way to solve for x, cause I'm having a hard time with this  :-\

Offline syko sykes

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Re: Help with Ksp
« Reply #1 on: May 22, 2006, 12:12:52 AM »
I got he equation 0=x2+0.1004-5.9 x 10-5. Now my question is, am I supposed to use quadratic formula or is there another way to solve for x, cause I'm having a hard time with this :-\
i don't think you have the right equation because solving that gives you an imaginary number for x

are you sure the equation is 0 = x2 + .1004 - 5.9E-5 ?
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Offline FouRRaW

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Re: Help with Ksp
« Reply #2 on: May 22, 2006, 12:22:10 AM »
I didn't read your question thoroughly but i know of this one thing my teacher show'd us so we don't have to use quadratic forumal. 

If you know that x is a very small number ex.(0.00001) as opposed to the other numbers being about 0.01 or something like that.  Use it too you advantage knowing that x is small, then you can add x + (any number) with ease.  Hope this helps

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Re: Help with Ksp
« Reply #3 on: May 22, 2006, 03:08:27 AM »
Please elaborate on the source of your equation and numbers used in it. In no way we can guess "the what of a what".
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Offline FouRRaW

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Re: Help with Ksp
« Reply #4 on: June 06, 2006, 11:27:10 PM »
x2                                    [H3O+] [A-]
                                                             [HA]
_____________

HA - x
                                                     HA + B = HB + A-


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