bball14 Posted November 21, 2004 Posted November 21, 2004 how could you find the pH for a .001 Molar solution of HCl? or what about .050 M of NaOH??
bball14 Posted November 21, 2004 Author Posted November 21, 2004 how could you find the pH for a .001 Molar solution of HCl? or what about .050 M of NaOH??
Gilded Posted November 21, 2004 Posted November 21, 2004 Remember that some compounds can donate more than one OH- or H+. H2SO4 is a good example; a diprotic acid, able to give up two protons. H3PO4 (phosphorous acid) is a triprotic acid, able to give three protons... and so on. That doesn't have anything to do with those HCl and NaOH of yours, but remember it anyway (if you didn't know it already ). Example: What about calculating pH when we use molar values? Well, the case is easy with H+ at least, since the amount of molecules in the solution doesn't actually increase, since the H+ stick to H2O making H3O+, a hydronium ion. 0.001mol HCl, 0.999mol H2O. We can use the 0.001 value, since that's also the molar value for the H3O+ (of course, then there's no HCl, just Cl- floating around). As we all know, or don't know, pH is -([H3O+]log), or the negative logarithm of the hydronium ion concentration. -(0.001log) = 3 You got yourself a pH 3 solution. This is where the diprotic-whatever-stuff comes in. If you had a solution with that same amount of water, and same amount of acid, but sulfuric acid in this case, you would get two protons from every H2SO4 molecule. -(0.002log) = approximately 2.7 That's about that. Does anyone know more about the pOH thing? Since I'm not perfectly sure if 0.1mol NaOH and 0.9mol water makes a 1.1 mol solution.
Gilded Posted November 21, 2004 Posted November 21, 2004 Remember that some compounds can donate more than one OH- or H+. H2SO4 is a good example; a diprotic acid, able to give up two protons. H3PO4 (phosphorous acid) is a triprotic acid, able to give three protons... and so on. That doesn't have anything to do with those HCl and NaOH of yours, but remember it anyway (if you didn't know it already ). Example: What about calculating pH when we use molar values? Well, the case is easy with H+ at least, since the amount of molecules in the solution doesn't actually increase, since the H+ stick to H2O making H3O+, a hydronium ion. 0.001mol HCl, 0.999mol H2O. We can use the 0.001 value, since that's also the molar value for the H3O+ (of course, then there's no HCl, just Cl- floating around). As we all know, or don't know, pH is -([H3O+]log), or the negative logarithm of the hydronium ion concentration. -(0.001log) = 3 You got yourself a pH 3 solution. This is where the diprotic-whatever-stuff comes in. If you had a solution with that same amount of water, and same amount of acid, but sulfuric acid in this case, you would get two protons from every H2SO4 molecule. -(0.002log) = approximately 2.7 That's about that. Does anyone know more about the pOH thing? Since I'm not perfectly sure if 0.1mol NaOH and 0.9mol water makes a 1.1 mol solution.
Guest TriggerDunpo Posted November 21, 2004 Posted November 21, 2004 the pOH is 14 - pH It's also the same things as pH, except just with the OH ions instead. Also, H2SO4 is a strong acid, so all of the first proton is lost, but HSO4- is a weak acid, so it forms an equilibrium in the solution. Not all of the second proton is lost to the solution.
Gilded Posted November 21, 2004 Posted November 21, 2004 "It's also the same things as pH, except just with the OH ions instead" Yes I know that, I was just wondering about the 0.1 mol NaOH 0.9 mol H2O case, since OH- doesn't have the same habit of sticking into H2O as H+ does. "HSO4- is a weak acid" Ahh, good point. Forgot about that. However, sulfuric acid is usually count as a diprotic acid (guess if it's just that it has the potential of being able to give up two protons).
chadn Posted November 22, 2004 Posted November 22, 2004 Yes I know that, I was just wondering about the 0.1 mol NaOH 0.9 mol H2O case, since OH- doesn't have the same habit of sticking into H2O as H+ does. It doesnt really matter, the principal is the same. The formula for each is exactly the same pH = -log [H+] and pOH = - log [OH-]. The fact that H+ forms H3O has no effect on the pH since both are the same thing. H+ and H3O are simply two different ways of representing the same thing. As long as we are refering to an aquaes solution.
Gilded Posted November 22, 2004 Posted November 22, 2004 "The fact that H+ forms H3O has no effect on the pH since both are the same thing. H+ and H3O are simply two different ways of representing the same thing." I know that, but because OH- doesn't stick to H2O, it's 0.1mol Na+, 0.9mol water and 0.1mol OH- in the example solution I presented. Isn't it? So does it count as a 1mol solution, telling that the OH- are virtually still attached to the Na+, or do you get 1.1 mol solution from this experiment?
chadn Posted November 29, 2004 Posted November 29, 2004 I know that, but because OH- doesn't stick to H2O, it's 0.1mol Na+, 0.9mol water and 0.1mol OH- in the example solution I presented. Isn't it? So does it count as a 1mol solution, telling that the OH- are virtually still attached to the Na+, or do you get 1.1 mol solution from this experiment? Im struggling to understand this, what your saying, but maybe this will help. Okay. Lets see. You cant go adding the 0.1 mol of Na+ and 0.1 mol of OH- and say you have 0.2 mol. It just doesnt work that way. Either you have 0.1 mol of each substance or you have 0.1 mol of NaOH.
budullewraagh Posted November 29, 2004 Posted November 29, 2004 that is correct: sum of mass of 0.1 mol Na+ and 0.1 mol OH-=mass 0.1 mol NaOH
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