how do i calculate [H] and [OH] CONCENTRATION USING THE Ph?
455471
pH (potential hydrogen) is a measure of the activity of hydrogen ions (H+) in a solution and, therefore, its acidity or alkalinity. In aqueous systems, the hydrogen ion activity is dictated by the dissociation constant of water (Kw = 1.011 × 10−14 at 25 °C) and interactions with other ions in solution. Due to this dissociation constant a neutral solution (hydrogen ion activity equals hydroxide ion activity) has a pH of approximately 7. Aqueous solutions with pH values lower than 7 are considered acidic, while pH values higher than 7 are considered alkaline..\n\npH = -log(concentration of H+)\njust use the equation to figure it out.\nhttp://en.wikipedia.org/wiki/PH
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
568277
Yeah, it can be done with just the line.  \nFirst, locate the end point on the line.  It's exactly the midpoint of the sharp pH rise or drop.  Next, find exactly half the end point, where the moles of salt equal the moles of whatever you're adding.  At that point, pKa = pH.
696206
the moles of NaOH = the moles of OH - = 0.125 * 0.606 = 0.07575\nthe moles of OH - in the new solution = 0.07575/(15 + 0.125 ) = 5*10^(-3) M\npOH = - log ( 5*10^ -3) = 2.3\n=> pH = 14 - pOH = 11.7
368556
The pH of a solution tells you the concentration of the H+ ions in solution.  However, if we know the pH, we can calculate the pOH and from that, the concentration of the OH- ions in solution.\n\npH + pOH = 14\nif pH = 8.25, pOH = 14-8.25 = 5.75\n\nJust like pH = -log([H+]),\npOH = -log([OH-])\n\nIf pOH = 5.75, that means that -log([OH-]) = 5.75, which mean that 10^-5.75 much give us the concentration of OH- ions.\n\n10^-5.75 = 1.778 E-6\nSo we know that the concentration of the OH- ions in the solution with a pH of 8.25 is 1.778 E-6 Molar.\n\n[OH-] = 1.778 E-6 M
952799
"p" is defined as minus the log...pH is saying -log(H).\n\nTherefore, -log(4) = 1 x 10 to the minus 4 = .0001\n\na pH of 8 = 1 x 10 to the minus 8 = .00000001\n\nso the change in concentration is 10 to the 4 or 10,000 times
863022
Here you are:\n\nFe+2 + 1/2O2 (g) + H2O (l)  ---> Fe (OH)2 (s)\n\nFe (OH)2 (s) + 1/2O2 (g) ---> Fe2O3 (s) + 2H2O (l)
410035
good news...10.63 is correct. Now how to get there.  \nMorp + H20 ---> Morp+ + OH-\ninitial concentration of Morp=.115M  everything else is 0.\nEquilibrium concentration of Morp=.115M-X, H20 ignore, Morp+=X, OH-=X.   Set up your equilibrium equation.  Kb=[OH-][Morp+]/[Morp]  pKb=-log Kb  so Kb=1.62e-6.  1.62e-6=[X][X]/[.115-X]  We can ignore the X subtracted from .115 since it is very small when compared.  so solving for x we finally get X=4.32e-4 which is the OH- concentration.  pOH=-log OH-  pOH = 3.36.  pH = 14-pOH .....pH=10.63.
727216
use beers law to figure out the concentration and then take that to mole ratio
255966
Use the hendeson-hasselbach equation\n\npH = pKa + log([A-]/[HA])\n\nand that you know the toatal concentration of ions\n.2 = [HA] + [A-]\n\nyou know the pKa (found in textbook or given) 4.18\nthe pH is 4.00\n\nso -.19 = log(A-/HA)  therefore .6457 = A-/HA\nand .2 = [HA] = [A-]\n\n... solve simultaneous equation\n\n[HA] = .122 and [A-] = .078\n\ndouble check\n\nK = ([H+][A-])/[HA]\n\nK = 10^-pKa = 6.45E-5\n[H+] = 10^-pH = 1E-4\n...and it does check out
894546
You did not put the volume of HCl to neutralize so we will do the neutralization to pH = 7 only.\n\nmoles of HCl = 35 0 g/l / 36.46 g/mole = 9.60 mol/l\nmoles of NaOH = 30 g/l / 40 g/mol = 0.75 mol/l\nOne mole of HCl will neutralize one mole of NaOH.\n\n9.60 * Vol(HCl) = 0.75 * V(NaOH).\n\nOnce you know if you have any NaOH left you can calculate how much NaOH to add to bring it to pH = 8 or 9\n\npH = 8 =====> pOH = 6  (pOH = 14 - pH)\nthen 6 = - log [OH]\nCalculate OH from last equation and this will give you the moles of NaOH you need to add to bring to pH = 8.  Same for pH = 9.
440505
pH = pKa + log ([Salt]/[Acid])\n\n[Salt] = 0.5 ; [Acid] = 0.75 ; ([Salt]/[Acid]) = 2/3\n\npKa = -log (1.85 x 10^-5) = 5-0.2671\n\nlog ([Salt]/[Acid]) = log 2/3 = -0.1761\n\nHence, pH = 5 - 0.2671 + (-0.1761)\n          = 5 - 0.4432\n          = 4.5568\n\nNote: All Calculations done using Clark's Tables. Use your calculator to calculate log for greater accuracy.
320061
Someone else has asked this same question before, and it really isn't that hard . . .\n\n60 ml of 0.250M HCl plus 60 ml of 0.125M NaOH will result in 120 ml of solution total.  All of the NaOH will have been neutralized by the HCl, leaving an excess of HCl, so the solution will be acidic -- the pH will be much less than 7.\n\n60 ml * (1.0 liter/1000 ml) * (0.250 mole/liter HCl) = .015 mole HCl\n\n60 ml * (1.0 liter/1000 ml) * (0.125 mole/liter NaOH) = .0075 moles NaOH\n\nThat means you will have 0.015 - 0.075 = 0.075 moles of HCl left over after the neutralization.\n\nI leave it for you to calculate the pH of a solution of 0.075 moles of HCl in 120 ml of solution.  Remember that pH = -log(base 10)[H+]
323248
pH = -log[H+]\n\nand\n\n[H+][OH-] = 1*10^-14\n\nwhere [ ] means concentration of.\n\n\nSo, to work out [H+], you divide 1*10^-14 by 3*10^-10 to get  3.33333*10^-5\n\nTake the -log of this, and you get 4.477.\n\nThe pH is 4.477.\n\nAll logs are in base 10.
850392
You have to use the methods for equilibrium and pH of a salt to calculate the concentration of salt that would produce that pH.\nI cannot find my notes from chemistry right now,\nhowever you need to know the definition of Ka(or Kb)\nthen set up the equation as the salt reacts.\n\nnh4+ + h20 <--> nh3 + h30+\n\npH = -log[H30+}, so your [h30+] = 10^-pH\nyou can then calculate the concentration of nh4 using molar ratios and such, and then you will have molarity.
864626
1) Calculate the initial concentration of the H+.\npH = - log [H+] = 2, therefore  [H+] = 10 ^ -2 mol / liter\n\n2) Calculate the dilution factor.\nWe want to change the concentration from 10 ^ -2 mol / liter to\n10 ^ -4 mol / liter. Therefore the dilution factor is 100.\n\n3) Calculate the amount of water to add.\nThe dilution factor is 100. We therefore want to have 100 x 14.5 ml = 1450 ml after dilution. We therefore have to add 1450 - 14.5 = 1435 ml of water
834164
Mg(OH)2  --->  Mg^2+  +  2OH^-\n\nk(sp) = [Mg^2+][OH^-]^2 \n\nIf pH = 12.8, then\npOH = 14 -12.8 = 1.2\ntherefore, [OH^-] = 10^1.2 =  0.0631 (ask your calculator!)\nnow, k(sp) = 1.2 * 10^-11 = [Mg^2+] [0.0631]^2 \nIf you solve for [Mg^2+] , you will have the answer.\n\nIf you need more help, please email me or re post.