Solution A has a pH of 5 and solution B has a pH Of 3. What is the hydrogen ion consentration in each solution
490327
solution A has a hydrogen ion concentration of 10^-5\nsolution B has a hydrogen ion concentration of 10^-3\nAll you have to do is 10^-ph\n\nIs this homework?
737009
a is true\nb is true\nc is not true\nd is true\ne is true\n\nBalanced equation is:\nCa(OH)2 + 2HCl ---> CaCl2 + 2H2O\n\nIonic equation is:\nCa(OH)2(s) + 2H3O(+)(aq) ---> Ca(+2)(aq) + 4H2O(l)
224213
well, pH = -log[H+]. so if you just undo the log you will have the concentration of H+ (which is the same as H30+). then .. [H+} + [OH-] = 14 good luck!
101093
I don't even get your question here... \n\npH is a mesurment of the quantity of Hydrogen Molecules that is available in a solution.  The pH of a solution changes depending on the concentration of salt you put in.
541583
pH from 0 to 7: Acid solution.\nPH from 7 to 14: Basic solution.
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<--------- 7 ---------->   (pH)\nacid                base\n\nthe smaller pH the stronger acid\nthe bigger pH the stronger base\n\nso you should arrange them from the smallest pH to the biggest\nstomach content (1.6)\nsoda pop (3)\nmilk (6.5)\nfresh water (7, of course neutral)\nhuman blood (7.4)\nsea water (8.7)\ndrain cleaner (8.7)
786992
b) small fraction of the maximum number of possible hydronium ions, H3O+.\n\nHNO2 + H2O <===> H3O+ + NO2-
249301
H3O+ .0292M = pH 1.53
172666
When a salt derived from a strong acid and a weak base dissolves in water, the solution becomes acidic. For example, consider the process\n \nNH4Cl(s) ----H2O ---> NH4+(aq) + Cl-(aq)\n\nThe Cl- ion has no affinity for H+. The ammonium ion NH4+ is  a weak conjugate acid of the base NH4+(aq) and ionizes as follows \n\nNH4+(aq) + H2O(l) <=> NH3(aq) + H3O+(aq)\n\nor simply  \n\nNH4+(aq) <=> NH3(aq)  + H+(aq)\n\nThe equilibrium constant ( or ionization constant) for this process is given by \n\n\n\nKa = Kw / kb  =   1.0*10^-14 / 1.8*10^-5 = 5.6*10^-10\n\n\n      x^2 / .10 =  5.6*10^-10\n\n           x^2  =   5.6*10^-11\n\n             x  =    7.5*10^-6 M  =  [H+]\n\npH  =  -log [H+]\n    \n    =  5.12\n\n     my answer is  C , NH4Cl
577084
Acid + base = salt + water.\n\nUnbalanced eqn is: Al(OH)3 + H2CO3 --> Al2(CO3)3 + H2O
1061977
(H+) + NH3 -> NH4+\n It makes an ammonium ion, if you are talking about a hydrogen ion and not H2.
766811
d. \n\nIt's a double replacement (acid/base) reaction.  The Sr and the H are both positive and they switch.  Since Sr has a 2+ charge and the PO4 has a 3- charge, you use a common multiple of 6.  The other product would be H(OH) or water.
679839
An acid has a higher concerntration of H+ ions in the solution. The PH scale is based on a logorithimic equation where a PH of 7 is neutral, andything below 7 is acidic and anything above 7 is basic. The conjugate acid is the H part of HX where the X- is the conjugate base. \n\nWikipedia has a great explination: http://en.wikipedia.org/wiki/Conjugate_acid
860365
use the equation  pH= -lgKa+ lg( salt/acid)\n(this equation only applies for buffer system)
62572
The concentration of hydrogen ion [H] and the concentration of hydroxide ion [OH] in any aqueous solution are related by an equalibrium expression, which is H2O in equalibrium with H+ and OH-.  There is an equalibrium constant associated with this Kw = 1.0 x 10^14 at 25 degrees Celcius.  [H] and [OH] are related.  Kw = [H]*[OH].  The simple and less technical answer is that as one concentration goes up say, [H], the concentration of the other [OH] goes down.
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HF + H2O => F + H3O+\nNH4+ + H2O => NH3 + H3O+\nH2CO3 + H2O => HCO3- + H3O+\nHNO2 + H2O => NO2 + H3O+\n\nbasically, the oxonium ion is formed in each one, when a proton is donated to water.
782516
a) 3SiO2 +2 Al --->3 Si + 2Al2O3\n\nb)AgNO3 +HCl---> AgCl +HNO3\n\nc) Mg + CuSO4 ---> MgSO4 + Cu\n\nd) Pb(NO3)4 +4 NaI --->4 NaNO3 + PbI4 (s)