Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: The Tao on May 18, 2007, 11:48:18 PM
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Ok the problem is
3.5g Ca(OH)2 + 500ml H2O -----> 2OH + Ca + H2O
1. Calculate pH (1.05)
2. Calculate pOH (0.74..?..)
Ok so I convert 3.5g Ca(OH)2 into .047mol, and then find it's molarity with .047mol/.5L (500ml) which is: [0.09]
Now I use -log(0.09) to find the pH, which = 1.05
Now what do I do to find the pOH? What I tried to do was use the balanced chemical equation to convert the molarity of Ca(OH)2 into the molarity of OH, and then use log. Is that correct?
[0.09]Ca(OH)2*2OH / 1Ca(OH)2 which gave me a molarity of [0.18]OH, and then I used -log(0.18)=0.74pOH
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1. How many OH- per Ca(OH)2?
2. pH + pOH = ? (hint: water ion product)
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Ok so I convert 3.5g Ca(OH)2 into .047mol, and then find it's molarity with .047mol/.5L (500ml) which is: [0.09]
Now I use -log(0.09) to find the pH, which = 1.05
This is pOH, not pH.
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Ok so I convert 3.5g Ca(OH)2 into .047mol, and then find it's molarity with .047mol/.5L (500ml) which is: [0.09]
Now I use -log(0.09) to find the pH, which = 1.05
This is pOH, not pH.
3.5g Ca(OH)2 + 500ml H2O -----> 2OH + Ca + H2O
By the way, is this reaction even correct? I wasn't given the reaction.
I'm kind-of confused why that's the pOH, but I'll believe you. So then if that's the pOH, can I still use the balanced chemical equation to convert .047mol Ca(OH)2 into mol OH? If so I get .094mol OH, and then .094molOH/.5L = [0.188], -log(.188) = pH= 0.72
1. How many OH- per Ca(OH)2?
2. pH + pOH = ? (hint: water ion product)
1. There are 2OH- for every Ca(OH)2. So I multiply by 2, and then find it's molarity with .5L and then find the pH?
2. I have no idea what the water ion product is. I don't think my teacher covered that.
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You're off by a multiplicative factor. You don't have 0.047 mol of HO-. You have twice as much HO- as Ca2+.
So this would mean you have how many moles of HO-?
Finding the concentration of HO- then, assuming 500 mL of water is roughly the solution volume, will be as you did before.
pH and pOH are p-functions, logarithmic functions of H+ and HO- concentration, respectively. So whatever your concentration of HO- came out to be, the p-function of that concentration would give you pOH, not pH.
To understand what the pH of the solution would be, then, you must employ a very common reaction that is true of all acid-base equilibria.
The relationship is commonly derived from the dissociation of water to H+ and HO- ions.
2H2O <---> H3O+ (source of H+) + HO-
If we designate HA as any acid, we can write an equilibrium for its dissocation in water (the same can be done for any base):
HA + H2O ---> H3O+ + A-
A- + H2O ---> HA + HO-
Now, applying your understanding of what an intermediate is (if not, review what an intermediate is), we can cancel the common intermediate(s) from both of these respective reactions, giving:
2H2O ---> H3O+ + HO-
Does this equation look familiar? So we can simply say that, according to this equation,
K = [H3O+][HO-]
[H2O]2
Since water is a pure liquid, it's concentration is a constant. If we bring the [H2O]2 term to the left, we get
K[H2O]2 = [H3O+][HO-], which can be re-written as
Kw = [H3O+][HO-]
This relationship can be written in other forms, notably using p-functionality:
pKw = pH + pOH
Do you see the significance?
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You're off by a multiplicative factor. You don't have 0.047 mol of HO-. You have twice as much HO- as Ca2+.
So this would mean you have how many moles of HO-?
Finding the concentration of HO- then, assuming 500 mL of water is roughly the solution volume, will be as you did before.
pH and pOH are p-functions, logarithmic functions of H+ and HO- concentration, respectively. So whatever your concentration of HO- came out to be, the p-function of that concentration would give you pOH, not pH.
To understand what the pH of the solution would be, then, you must employ a very common reaction that is true of all acid-base equilibria.
The relationship is commonly derived from the dissociation of water to H+ and HO- ions.
2H2O <---> H3O+ (source of H+) + HO-
If we designate HA as any acid, we can write an equilibrium for its dissocation in water (the same can be done for any base):
HA + H2O ---> H3O+ + A-
A- + H2O ---> HA + HO-
Now, applying your understanding of what an intermediate is (if not, review what an intermediate is), we can cancel the common intermediate(s) from both of these respective equilibria, giving:
2H2O ---> H3O+ + HO-
Does this equation look familiar? So we can simply say that, according to this equation,
K = [H3O+][HO-]
[H2O]2
Since water is a pure liquid, it's concentration is a constant. If we bring the [H2O]2 term to the left, we get
K[H2O]2 = [H3O+][HO-], which can be re-written as
Kw = [H3O+][HO-]
This relationship can be written in other forms, notably using p-functionality:
pKw = pH + pOH
Do you see the significance?
But, I didnt' say that I had .047 mol HO- I said I had .047 mol Ca(OH)2. I then used the balanced equation to covert .047 mol Ca(OH)2 into .094mol HO-, which IS twice as much Ca(OH)2. Then I divded .094mol HO- by .5L (500ml) to get it's molarity, which is [.188] and then finding the pH using -log(.188) = 0.72
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I have no idea what the water ion product is. I don't think my teacher covered that.
g_english explained details (you may check my water ion product (http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product) page as well). However, you can use Google to check what water ion product is even if you don't know. In real life when you don't know something nobody will ask you if your teacher covered that or not - you will just get fired.
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Your pOH is 0.73, not your pH.
Think about it, you have a mineral base (mineral bases are all strong bases).
How can you justify this solution having an acidic pH (a low pH, according to your answer)?
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Ok. Let me try to walk myself through this again after researching for a bit.
3.5g Ca(OH)2 is dissolved in 500ml H2O. Calc. pH and pOH
I write the balanced chemical equation out.
Ca(OH)2 + H2O ---> Ca + 2OH- + H2O
Now I convert 3.5g Ca(OH)2 into .047 mol Ca(OH)2. I use the .047mol Ca(OH)2 to convert it into 2OH- which is .094mol OH-. And then I use .094mol OH-/.5L = OH-[.188] pOH = -log(.188)=
Do I then use the equation [H+][OH-]=1.0x10-14 to find the [H+], and THEN plug that back into -log(H+) to find the pH?
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Yes.
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But then, [OH-=[.188] -log(.188) = pOH= 0.72
Isn't that value acidic?
Then pH = 13.2, isn't that basic? lol...
Should they be switched around? If so, where did I make my mistake in the math?
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But then, [OH-=[.188] -log(.188) = pOH= 0.72
Isn't that value acidic?
No. No math error.
You're confusing the meaning of pOH with pH. A low pH value indicates an acidic solution. A low pOH value represents a basic solution.
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It's funny that my book conviently left that part out.
Anyways...then this question would be bettered described as finding the pH and pOH of a strong base?
Then let me ask another question here:
A. Calculate the pH of a [0.04] HF solution. Ka=4.0x10-4
B. Calculate [F-]
I write out the balanced chemical equation: HF + H2O ---> F- + H3O and everything seems to be at a 1/1 ratio.
For A, I use the equation [H+]= sqroot of (Ka)([M]), aftering plugging the values in I get,
[H+]= sqroot of (4.0x10-4)([0.04])
Which equals = [H+]= 4.0x10-3 and then pH= -log(4.0x10-3) = 2.39
For B, I just use the equation (4.0x10-3)(F-)= 1.0x10-14 or is this a differen't value?
In my notes I show that I didn't use this equation, but that I used HF + H2O ---> 4.0x10-3 F- + 4.0x10-3 H3O since it's a 1/1 ratio. Is my first attempt or this one correct for B?
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pH and pOH, no superscripts.
sqroot of (Ka)([M]) works only if the dissociation fraction is below 5%, you have to check it before using [H+] calculated this way, as you may have to use more complicated equation.
See pH of weak acid (http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base) calculation.
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Well. sqroot of (Ka)([M]) and Ka=[H+][OA-]/[HoA] are the only equations he showed us regarding weak acids. Can I solve this problems using only these two equations?
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Dissociation constant is the main one, this plus stoichiometry of dissociation are almost always enough to perform calculations. Still, you need to know what and how. That's described on the page I have pointed to, with a shortcut (in the form of pH cheat sheet) linked at he very bottom of th electure - but don't use it till you will get the basics.
The square root equation can be derived with some sipmilfying assumptions from the dissociation constant; see equation 8.13.