how do i calculate [H] and [OH] CONCENTRATION USING THE Ph?
404745
The "p" stands for negative log.\nSo, the negative log of the concentration of H+ ions in solution gives you the pH.  \n-log[H+] = 3.40\nMultiply both sides by negative one:\nlog[H+] = -3.40\nNow raise 10 to the both sides.  This will cancel out the log function. \n10^log[H+] = 10^-3.40\n[H+] = 3.98E-4 M\n\nTo get the [OH-], just divide the [H+] into 1.00E-14:\n1.00E-14 / 3.98E-4 = 2.51E-11 M
639844
[H3O+]=1 x 10^-4 => pH=-log[H+]=4 --- acidic\n[OH-]=1X10^-6 => pOH=-log[OH-]=6 => pH=14-6=8 --- basic\n\nHCl --> H+ + Cl-\n[H+]= 5.6 x 10^-4\npH = -log[H+] = -log[5.6 x 10^-4] = 3.25 --- acidic\npOH = 14 - 3.25 = 10.75\n-log[OH-] = 10.75 => [OH-]=1.78 x 10^-11\n\nSince HBr --> H+ + Br-\npH of 1 x 10^-4 of HBr = -log[1x10^-4] = 4 --- acidic\n\nSince Ca(OH)2 --> Ca2+ + 2OH-\npOH = -log[OH-] = -log(2 * 5 x 10^-4) = 3\npH = 14-3 = 11 --- basic
553835
Well to determine [H+] you use the equation that \npH=-log [H+] \nthus a solution with pH=3 will have a [H+] of 1x10^-3, \nand a solution with pH=8 will have a [H+] of 1x10^-8\nTo find [OH-] there are several equations you can use\n[OH-]=Kw/[H+], \nKw is the ionization constant of water which is equal to 1.0x10^-14\nThus you will have [OH-] of 1.0x10^-11 and 1.0x10^-6\nSo you answer will be that the solution with pH=3 will have a [H+]>[OH-]
249301
H3O+ .0292M = pH 1.53
620217
HOCl <==========>  H+  + OCl-\n\nthe pKa of hypochlorous acid, HClO, is 7.4 at 25°C, so:\n\npKa = -log [Ka] = 7.4, therefore\n\nKa = 3.98 x 10^-8\n\nKa = [H+]*[OCl-]/[HOCl] = 3.98 x 10^-8\n\nfrom your problem, you state that you have 1.0M HOCl, and since the only source for ions is the HOCl dissociating;\n\n[H+] = [OCl-] = y, and [HOCl] = 1.0-y\n\nsubstituting this back into the Ka equation, we get:\n\n3.98 x 10^-8 = y^2/(1.0-y)\n\nwhich converts into the quadratic equation:\n\n0 = y^2 + (3.98x10^-8)y - 3.98x10^-8\n\nI leave it to you to solve the quadratic equation to get y = [H+], and then calculate the pH of the starting solution
757438
CaC2 + 2 H2O ---> Ca(OH)2 + HCCH
438792
Okay. This is a multi-step pH problem. First, find the Kb of the fluoride ion (F-). To do this, follow this formula:\n\nKw/Ka = Kb\n\nThis formula can be rearranged to be Kw/Kb = Ka, by the way. Kw is the K of dissociation of water, which is 1.0e-14.\n\nPlug in the numbers and you get 1.47e-11 as a Kb for fluoride. Now, write an equilibrium ICE table, filling in the values you know.\n\n_F- + H2O --> HF + OH-\nI\nC\nE\n\nSolve the equilbrium table and you get the following equation:\n\nx^2/(0.00339-x) = 1.47e-11\n\nx is the concentration of both HF and OH-.\n\nSolve the equation and you get 2.23e-7 as the concentration of OH-. The easiest way to finish the problem now is to find the pOH. Take the -log of the concentration of OH-. The pOH is 6.651. Now, follow this equation: pH + pOH = 14. The pH of the solution is 7.34876. To the right amount of significant figures, the answer is 7.349 (the numbers to the left of the decimal place doesn't count toward the sig figs in pH's).\n\nHope this helped.
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.
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.
781593
I always have to derive the equations from scratch:\n\nKa = [H+][A-]/[HA]\n\nlog(Ka) = log(H+) + log(A-) - log(HA), multiple both sides by -1\n\npKa = pH + log(HA) - log(A)\n\npH = pKa + log(A-/HA)\n\n4.54 = 4.74 + log(A-/HA)\n\n-0.20 = log(A-/HA)\n\n10^(-0.20) = A-/HA = 0.631\n\nplease check for errors
592167
We assume NaOH is a strong base and therefore dissociates completely at 60 degrees C. a concentration of .002M solution of NaOH dissociating completely contributes .002 mols/liter of OH ions to the solution. Since pOH = -log[concentration of OH-], pOH = -log.002 = 2.699 (with correct logarithmic sig figs). pH is 14 - pOH, so the pH of the solution will be 14-2.699 = 11.301.
606155
NaOH + HCl ---> NaCl + H20
901261
I think that the equation is \n3Ba(OH)2 + 2H3PO4 ---> Ba3(PO4)2 + 6H2O\nbut maybe the answer is\nBa(OH)2 + 2H3PO4 ---> Ba(H2PO4)2 + 2H2O\nbut no problem you can use the same  way to calculate.\n\n3Ba(OH)2 + 2H3PO4 ---> Ba3(PO4)2 + 6H2O\nthe moles of H3PO4 0.01015 * 0.141 = 1.43 * 10^-3\nthe moles of Ba(OH)2 needed : 1.43 * 10^-3 *3 / 2 = 2.145 * 10^-3\nthe volume of Ba(OH)2 needed : 2.145 * 10^-3 / 0.0584 = 0..03673 L = 36.73 ml
795289
If the solution contains a diprotic acid, there are two hydrogen atoms available for the acid to lose (such as H2SO4).\n\nA solution of 50mL of a 1.012 M solution would contain 0.0506mol of diprotic acid which would then dissociate to give you 0.1012mol of H+ ions. \n\nThen you must have as much hydroxyl anions (OH-) as you do hydrogen ions (H+), so you must have 0.1012mol of OH-. Using the formula 2H+ + 2OH- -----> 2H2O.\n\nTo convert 0.1012mol OH- into grams of Mg(OH)2, you divide the OH- concentration in half because of the equation\nMg2+ + 2OH- -----> Mg(OH)2\n\nwhich would give you the 0.0506mol Mg(OH)2.\n\nFinally you would convert the 0.0506mol Mg(OH)2 into grams Mg(OH)2 by multiplying the number of mol by the molecular weight of magnesium (II) hydroxide--the calculation is shown below:\n\n0.0506mol x 58.31g/mol = 2.95 g Mg(OH)2\n\nSo you would need 2.95 grams of Mg(OH)2 to neutralize the solution. \n\nHope this helps!!
968497
The initial concentration of the acid is C=mole/V=1/2=0.5 mole/L\nLet's assume that the acid is HA\n\n.. .. .. .. .. .. .. HA <=> H(+) + A(-)\nInitial .. .. .. .. 0.5\nDissociate .. .x\nProduce .. .. .. .. .. .. .. x .. .. .. x\nAt Equil. .. . 0.5-x .. .. . x .. .. .. x\n\nKa=[H+][A-]/[HA] =x^2/(0.5-x)\n\nbut x=[H+]=3.4*10^-2 so\n\nKa= ((3.4*10^-2)^2) / (0.5-3.4*10^-2) =2.48*10^-3
427689
seems like a homework question so use the equation pH=pKa + log [base]/[acid].  As a check for your answer when [base]=[acid]  the pH will equal the pKa.  the acid and base are  close in concentration so your answer will be around a pH=9.  pOH = 14-pH will help you on the other question.
407734
Hummm:  You can calculate the [OH-] from the pH.  [OH-] = 10^-2.  pH + pOH = 14 (from [H+]x[OH-] = 10^(-14):  pOH = 14 - 12 = 2.  Find the solubility product for iron (III) hydroxide in a table in the back of your textbook or a handbook of Chemistry or physics. Kp = [Fe+3] x [OH-]^3. Substitute numbers in and solve for [Fe+3].  This is also the molar solubility as one mole of iron hydroxide contains one mole of Fe+3.