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If I were to find the pH of a solution of Ba(OH)2, am I supposed to multipy the log of the molarity by 2, since there are 2 OH- ions?
Isn't Ba(OH)2 only partially soluble? So once you use ksp to get [OH-] just get the log of that, and subtract it from 14.....If I were to find the pH of a solution of Ba(OH)2, am I supposed to multipy the log of the molarity by 2, since there are 2 OH- ions?
Isn't Ba(OH)2 only partially soluble? So once you use ksp to get [OH-] just get the log of that, and subtract it from 14.....
Or using a numerical example, lets say pH is 9, so pOH is 5
[OH-]=1*10^-5
Not sure what you're asking exactly.....
Wouldn't multiplying the log by 2 mean that you're squaring the [OH-]?Ba(OH)2 is one of the soluble metal hydroxides (along with Sr(OH2) and Ca(OH2) ). Yeah, I thought that's how it's usually done, take the negative log, then substract from 14 but my friend said something about multiplying the concentration or the log of the concentration by 2 to account for the 2OH- ions or something? Seemed weird to me, as I've never heard of that before, and so I just wanted to double check here...
Wouldn't multiplying the log by 2 mean that you're squaring the [OH-]?
What you can do is mutliply the molarity of Ba(OH)2 as long as it's in the range before it gets saturated.
So if [Ba(OH)2] = 1*10^-5 then [OH-] = 2*10^-5 and pOH = 4.X.....
Ok so say [Ba(OH)2]=5*10^(-5). To get the concentration of OH-, you'd have to multiply by 2. Because when it dissociates you get 2 moles of OH- for each mole of Ba(OH)2. So [OH-]=10*10^(-5)=1*10^(-4). So your pOH is 4 and the pH is 10.
Make sense?
Yeah, that makes sense. So if you were to find the concentration of Ba(OH)2, and it was 1X10^-5, then just the OH concentration would be 2X10^-5?
Would this be the case if they told you the PH was 9? b/c POH would b 5. Then [OH]- would be 10^-5, like you said before, right? So maybe it wouldn't? Ahh >.<