Question about pH
Started by G1SG2
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.....
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...
Yea, can anyone else chime in on this? Now that I've read your question I'm confused as well. Do we just do it as normal, or is there another step involved? I would have just done it like normal if I saw a similar problem a min ago, but now that I look at it that way, it makes sense to multiply by 2 (sort of?). Aahh, the MCAT will be the death of me.
Advertisement - Members don't see this ad
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.....
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 >.<
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 >.<
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?
Make sense?
Yes, this is right. pH + pOH = 14. Always. And pOH = -log[OH-]. Always.
If they told you the pH was 9 and asked for the original concentration of Ba(OH)2, you just go backwards. pOH=5 so [OH-]=1*10^(-5). You need two moles of OH- per mole of Ba(OH)2 so your original concentration of Ba(OH)2 = .5*10^(-5)=5*10^(-6).
Is this kinda what you are getting at? I'm not sure.
