G1SG2

10+ Year Member
5+ Year Member
May 2, 2008
1,454
2
Status
  1. 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?
 

wanderer

10+ Year Member
5+ Year Member
Dec 14, 2008
1,979
30
Status
  1. Medical Student
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.....
 

G1SG2

10+ Year Member
5+ Year Member
May 2, 2008
1,454
2
Status
  1. Pre-Medical
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...
 
About the Ads

vandyam

Vulnera Sanentur
5+ Year Member
Jun 17, 2009
483
2
Dirty South
Status
  1. Medical Student
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.
 

wanderer

10+ Year Member
5+ Year Member
Dec 14, 2008
1,979
30
Status
  1. Medical Student
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.....
 

G1SG2

10+ Year Member
5+ Year Member
May 2, 2008
1,454
2
Status
  1. Pre-Medical
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.....

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 :oops:
 

vandyam

Vulnera Sanentur
5+ Year Member
Jun 17, 2009
483
2
Dirty South
Status
  1. Medical Student
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 >.<
 

ksmi117

GEAUX TIGERS!!!
Moderator Emeritus
Lifetime Donor
10+ Year Member
Mar 16, 2008
21,960
179
Status
  1. Resident [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?
 

G1SG2

10+ Year Member
5+ Year Member
May 2, 2008
1,454
2
Status
  1. Pre-Medical
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?

Yesss! Thanks :)
 

ksmi117

GEAUX TIGERS!!!
Moderator Emeritus
Lifetime Donor
10+ Year Member
Mar 16, 2008
21,960
179
Status
  1. Resident [Any Field]
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 >.<

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.
 

vandyam

Vulnera Sanentur
5+ Year Member
Jun 17, 2009
483
2
Dirty South
Status
  1. Medical Student
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 :D
 
This thread is more than 11 years old.

Your message may be considered spam for the following reasons:

  1. Your new thread title is very short, and likely is unhelpful.
  2. Your reply is very short and likely does not add anything to the thread.
  3. Your reply is very long and likely does not add anything to the thread.
  4. It is very likely that it does not need any further discussion and thus bumping it serves no purpose.
  5. Your message is mostly quotes or spoilers.
  6. Your reply has occurred very quickly after a previous reply and likely does not add anything to the thread.
  7. This thread is locked.
About the Ads