- Joined
- May 2, 2008
- Messages
- 1,454
- Reaction score
- 2
- Points
- 4,571
- Pre-Medical
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?
Question about pH
- Thread starter G1SG2
- Start date
- Joined
- Dec 15, 2008
- Messages
- 1,979
- Reaction score
- 33
- Points
- 4,641
- Medical Student
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.....
- Joined
- May 2, 2008
- Messages
- 1,454
- Reaction score
- 2
- Points
- 4,571
- Pre-Medical
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.
- Joined
- Dec 15, 2008
- Messages
- 1,979
- Reaction score
- 33
- Points
- 4,641
- Medical Student
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.....
- Joined
- May 2, 2008
- Messages
- 1,454
- Reaction score
- 2
- Points
- 4,571
- Pre-Medical
Good point. I suppose then, it wouldn't make that much of a difference in the calculation? Sorry if I'm confusing anyone...this confused me too, lol 😳
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 >.<
- Joined
- Mar 16, 2008
- Messages
- 21,964
- Reaction score
- 181
- Points
- 5,266
- Fellow [Any Field]
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?
- Joined
- Mar 16, 2008
- Messages
- 21,964
- Reaction score
- 181
- Points
- 5,266
- Fellow [Any Field]
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.
Yeah, this was great thanks. I just wasn't separating the two ideas: The concentration of Ba(OH)2 isn't necessarily equal to the [OH-] concentration. Got it -- thanks a lot 😀
