Quantcast

Question about pH

Get Shadowing and a Virtual Clinical Education
This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

G1SG2

Full Member
10+ Year Member
5+ Year Member
Joined
May 2, 2008
Messages
1,454
Reaction score
2
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

Full Member
10+ Year Member
5+ Year Member
Joined
Dec 15, 2008
Messages
1,979
Reaction score
32
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

Full Member
10+ Year Member
5+ Year Member
Joined
May 2, 2008
Messages
1,454
Reaction score
2
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...
 

vandyam

Vulnera Sanentur
5+ Year Member
Joined
Jun 17, 2009
Messages
483
Reaction score
2
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

Full Member
10+ Year Member
5+ Year Member
Joined
Dec 15, 2008
Messages
1,979
Reaction score
32
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

Full Member
10+ Year Member
5+ Year Member
Joined
May 2, 2008
Messages
1,454
Reaction score
2
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
Joined
Jun 17, 2009
Messages
483
Reaction score
2
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
Joined
Mar 16, 2008
Messages
21,964
Reaction score
181
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

Full Member
10+ Year Member
5+ Year Member
Joined
May 2, 2008
Messages
1,454
Reaction score
2
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
Joined
Mar 16, 2008
Messages
21,964
Reaction score
181
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
Joined
Jun 17, 2009
Messages
483
Reaction score
2
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
 
Top