2006-12-16 22:43:22 · 5 answers · asked by Yuvraj K 1 in Science & Mathematics ➔ Chemistry
Normality and Molarity are connected through the equation N=aM
H2SO4 has 2 H+, thus a=2
The good thing about normality is that it shows you how much is the concentration of the chemical equivalent units. In this case the units are H+, so the normality of the acid actually corresponds to the molar concentration of H+ provided by the acid. (You would reach the same conclusion if you used the above equation to convert acid normality to molarity and then find out the [H+]).
So 36 N acid (of any acid with 100% dissociation) corresponds to [H+]=36 M. This is a very concentrated solution so you wouldn't express its acidity with the pH scale. If you want to do it, pH=-log[H+]= -log(36) = -1.56
10^-15 N acid corresponds to [H+]=10^-15 M. This is extremely dilute. The self-dissociation of water is much more significant than this so the pH will be practically 7. If you want an exact value, then you have to consider the self dissociation of water in the presence of the H+ coming from the acid
Kw=[H+][OH-]= (10^-15+x)x =10^-14 =>
x^2+ 10^-15x - 10^-14=0 =>
x=10^-7 ...
and pH=-log(10^-15+10^-7) = 6.9999999957= 7.00 as you would expect for pure water.
This approach is meaningful only when the H+ coming from the acid is in the same order of magnitude as that coming from water (strong acid concentration range 10^-6 - 10^-8 or maybe 10^-9 M)
2006-12-17 00:45:24 · answer #1 · answered by bellerophon 6 · 0⤊ 0⤋
The pH of such a concentrated acid as 36N (I believe that is the same as normal concentrated sulphuric acid at 18M) is meaningless. It does not have enough water to ionise very much at all. The best answer would be 7.0.
Similarly, the very dilute one will have a pH of 7.0. There is not enough acid present to shift the equilbrium H+ + OH- <---> H2O at all.
2006-12-17 01:42:37 · answer #2 · answered by Gervald F 7 · 0⤊ 0⤋
heyheyfuff, For sulfuric acid, there are actual 2 feasible dissociations: H2SO4 ---> H^+ + HSO4^- HSO4^- ---> H^+ + SO4^2- both reactions have their personal dissociation constants : Ka(a million) = [HSO4^-]*[H^+] / [H2SO4] Ka(2) = [SO4^2-]*[H+]/[HSO4^-] that is known that the first proton comes off thoroughly (it extremely is, its Ka(a million) is sufficiently vast no longer to be a element -- it extremely is why they in effortless words gave you the 2d Ka for HSO4^-. assume (by means of the undeniable fact that first dissociation is finished) that we've a million.30*10^-3 M HSO4^-, then enable it dissociate: ***HSO4^- ---> H^+ + SO4^2- ***Ka(2) = [SO4^2-]*[H+]/[HSO4^-] = a million.20*10^-2 for each x moles/liter that dissociates, [HSO4^-] decreases by technique of x = (a million.30*10^-3 - x) [SO4^2-] will advance by technique of x = x [H+] will advance by technique of x = x Plug those into the equation for Ka(2), and sparkling up for x: Ka(2) = a million.20*10^-2 = (x)*(x)/(a million.30*10^-3 - x) (x^2) = (a million.20*10^-2)*(a million.30*10^-3) - a million.20*10^-2*x x^2 + (a million.20*10^-2)x - a million.fifty six*10^-5 = 0 sparkling up the quadratic equation, and the in effortless words rational answer is 0.00118, or a million.18 x 10^-3M. The [H+] = (a million.18 x 10^-3), so pH = -log[H+] = -log(a million.18 x 10^-3) = 2.ninety 3 desire that helped!
2016-11-26 23:58:59 · answer #3 · answered by ? 4 · 0⤊ 0⤋
what are 36N and 10^-15N
I must know that to solve the problem for you
2006-12-16 23:09:11 · answer #4 · answered by James Chan 4 · 0⤊ 0⤋
36 normal is very concentrated
therefore very close to 1
2006-12-17 00:45:34 · answer #5 · answered by b r 4 · 0⤊ 0⤋