Posted by Henry on Saturday, August 6, 2011 at 3:55pm.
a shampoo solutino at 25 C has a hydroxide ion concentration of 1.5 X a0^-9 M. Is the solutino acidic, neutral, or base
Henry, what don't you get about this. It's all about (H^+)(OH^-) = Kw = 1E-14 and pH = -log(H^+).
Sr(OH)2 is 0.005M so OH^- must be twice that. Go from there.
can you tell me whats the difference between hydronium and hydroxide ions are??
in my book the examples only says H+ ions
do you mind solving one of them for me? im kinda confused
Henry. Hydronium is the positive ion, shortened to H+ in your book. Hydroxide is the OH- ions.
because [H+][OH-] in water solutions is Kw (above), if you know one, you can solve the other. { [] means concentration in molarity)
Here, it is given that Sr(OH)2 concentration is .005M, so OH- concentration must be .010M
[H+][OH-]=1E-14
[H+]=1E-14/.010= you do it.
Now, if [H+] is greater than 1E-7, it is acidic
remember pH= -log [H+]
It bothers me you are confused on these. Do you need a regular tutor? If so, get on quickly, do not tarry.
just to see if im on the right track is the answer for the first equation
[H+]=Kw/[HCl]=(1.0*10^-14)/0.020=5.0*10^-13
??
^^ second equation*
H^+ and H3O^+ (hydrogen ions and hydronium ions) are used in many texts as the same thing. Although they are slightly different [(H3O^+) is just a hydrated H^+], most people who use them use them as being the same. At least for problems they can be considered the same.
Hydroxide ions, on the other hand, are OH^-. When H2O ionizes it does so to produce H2O ==> H^+ + OH^- (a hydrogen ion and a hydroxide ion) and the ion product of (H^+)(OH^-) = a constant which at about 20 C is 1E-14. Therefore, if you know one of them you can calculate the other one. [Note: if we want to be more precise about this the ionization equation is this,
H2O + H2O (which I will write to help you see it as H2O + HOH ==> H3O^+ + OH^- and (H3O^+)(OH^-) = Kw = 1E-14.
Now for the Sr(OH)2 problem. Sr(OH)2 is a strong base; therefore, it ionizes 100%. Sr(OH)2 ==> Sr^2+ + 2OH^-
So if Sr(OH)2 is 0.005, then Sr^2+ = 0.005 and OH^- is 2*0.005 = 0.01.
So (H^+)(OH^-) = 1E-14
(H^+)= 1E-14/(OH^-) = 1E-14/(0.01) = 1E-12. Since pH = -log(H^+), then pH = -log(1E-12) = -(-12) = +12.
There are other ways of doing this which I use most of the time because I thinks its faster. Since pH = -log(H^+), it follows that pOH = -log(OH^-) {and for future reference it also follows that pKa = -log Ka).
So if (H^+)(OH^+) = Kw then
pH + pOH = pKw = 14. So I do this in the problem.
OH^- = 0.005*2 = 0.01.
pOH = -log(0.01) = -(-2) = 2
pH + pOH = 14 so
pH + 2 = 14 and pH = 12
On the pH scale, pH = 7 is neutral
pH>7 = basic
pH<7 = acidic.
For the HCl problem, HCl is a strong acid. It has no ionization constant.
HCl ==>H^+ + Cl^-
If HCl is 0.20 M then H^+ = 0.20 M
OH^- = (1E-14/0.2) = ?? but I don't think its 5E-13.(I think you used 0.02 and not 0.20.)
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