This is a good point. In the lab, when we make solutions we talk about the concentration of a compound; this is just how it's done by convention. We just have to remember that in any strong electrolyte, the species present will actually be ions.
Thank you so much for the video!😀 I'm bit confuse about the calculation of [H+] and [OH-] in Ca(OH)2 dissociation, why use the equation of water dissociation([H+][OH-]=10^-14) to calculate the [H+]? (shouldn't this equation only used for water dissociation? I mean in this scenario, the Ca(OH)2 added, [OH-] should be larger than [OH-] produced by water dissociation, and [H+] won't change, so the entire concentration product of H+ and OH- should be larger than 10^-14? if the water dissociation can be ignored in this scenario, then where does [H+] come from? Shouldn't we add the [H+] and [OH-] produced by water dissociation?)
Why we find Molarity of the final solution? We need total [H+] for the formula PH=-log[H+], Can’t we just plug in total [H+], which is .1100 in the formula for PH=-log [H+] and disregard finding total Molarity instead?
.1100 is moles, not molarity. The pH is the negative log of the hydrogen/hydronium ion concentration, not moles! The brackets around the H+ in the equation indicate the molarity of H+. We have to divide the total moles by the total liters to find molarity because we can only plug in that value to make the equation work. On the other hand, plugging in moles for the [H+] will give you a wrong answer. Sorry for the late explanation, but I hope this helps! :)
I understand conceptually why a pH of 1.0*10^-14 M of HCl would have a pH of 7 because of it being so dilute but then why does the -log[H+] equation not work for this scenario??
This is because at this dilute level, the vast majority of H+ ions come from the autoionization of water. Since that is around 1.0 x 10^-7 at 25°C, the pH would be right around 7.00.
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thank you so much !! your explanations are so clear to me, because of you i’m feeling more confident hopefully it’ll translate on the exam :)
Thanks for your kind words. My goal is to make my explanations as clear as possible. I hope you do great on the exam!
At about 5 minutes, a question asks about the concentration of Ca(OH)2 (aq). But how can it have concentration if it completely dissolves into ions?
This is a good point. In the lab, when we make solutions we talk about the concentration of a compound; this is just how it's done by convention. We just have to remember that in any strong electrolyte, the species present will actually be ions.
Thank you so much for the video!😀
I'm bit confuse about the calculation of [H+] and [OH-] in Ca(OH)2 dissociation,
why use the equation of water dissociation([H+][OH-]=10^-14) to calculate the [H+]?
(shouldn't this equation only used for water dissociation?
I mean in this scenario, the Ca(OH)2 added, [OH-] should be larger than [OH-] produced by water dissociation, and [H+] won't change, so the entire concentration product of H+ and OH- should be larger than 10^-14?
if the water dissociation can be ignored in this scenario, then where does [H+] come from?
Shouldn't we add the [H+] and [OH-] produced by water dissociation?)
What about sulfuric acid in which there are 2 H's per mole? Wouldn't we double molarity of sulfuric acid to find H+ concentration
Not exactly, because only one of the two H's in H2SO4 is strong. The dissociation will be H2SO4 --> H+ + HSO4-. That second H is significantly weaker.
Why we find Molarity of the final solution? We need total [H+] for the formula PH=-log[H+], Can’t we just plug in total [H+], which is .1100 in the formula for PH=-log [H+] and disregard finding total Molarity instead?
.1100 is moles, not molarity. The pH is the negative log of the hydrogen/hydronium ion concentration, not moles! The brackets around the H+ in the equation indicate the molarity of H+. We have to divide the total moles by the total liters to find molarity because we can only plug in that value to make the equation work. On the other hand, plugging in moles for the [H+] will give you a wrong answer. Sorry for the late explanation, but I hope this helps! :)
I understand conceptually why a pH of 1.0*10^-14 M of HCl would have a pH of 7 because of it being so dilute but then why does the -log[H+] equation not work for this scenario??
This is because at this dilute level, the vast majority of H+ ions come from the autoionization of water. Since that is around 1.0 x 10^-7 at 25°C, the pH would be right around 7.00.
@@JeremyKrug ahh ok, thank you so much