the pH of a solution is 8.25, what is the (OH-)?
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
939677
Long answer:  Use the vol. and concentration of NaOH to figure out how many moles of NaOH you have which will = moles of OH (since NaOH has only one OH).  Formula is:\n\nMolarity = moles/liters (make sure the volume is in liters)\n\nIn a neutralization reaction the moles of OH = moles of H+\n\nOnce you have moles of H+, use that answer and the volume of HCl to calculate concentration of H+ which will be = to concentration of HCl since HCl has only one H+\n\nUse the same Molarity formula\n\nShort answer:  As long as the base (NaOH) and the acid (HCl) have only one OH and one H, the short form is:\n\nMb x Vb = Ma x Va\n\nwhere Mb = molarity of the base\nwhere Vb = volume of the base\nwhere Ma = molarity of the acid\nwhere Va = volume of the acid
853922
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
574036
The first step, find the moles of both.\nYou should get 25 mmol benzoic acid, and 15.6 mmol NaOH.\n\nThen you must find after the reactions, how many moles remain.  \nYou should get 9.4 mmol benzoic acid, 0 moles NaOH, and 15.4 mmol benzoate.\n\nYou have some benzoic acid and some benzoate ion, a conjugate acid/base pair.  This is a buffer solution, so use the Henderson-Hesselbach equation.  You should get pH = 4.4.
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.
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
1118024
Acidic \n\nbelow 6 acidic above 7 basic
818293
im pretty sure but  not positive\nC3H8 + 5 O2  -->  4 H2O + 3 CO2\n\nso 5 mol O2 to 1mol C3H8
1073643
the easiest way is to use a pH paper to test the pH. and just compare the reading with the one given with the test kit. \n\nOr you may do some neutralisation. And then calculate with the  amount of alkali you use for neutralisation. By that you can calculate the concentration of hydroxide ions in the alkali that you use. Then calculate the pOH for the alkali you use. pH=14 - pOH.\neg. \n1mol of NaOH gives you 1 mole of OH ions.\nso the pOH = -log1 = 0\npH = 14 - 0 = 14\n\nJust some rough guide. Hopefully it helps...Good luck.
745984
To answer this question, you must know the Ka or Kb values for those acids or bases above.\nYou must then use those values to find the concentration of H+ (H30+ depending how you were taught it) or OH- ions.\nthese values can then be used to find pH or pOH depending which ion value you have.\nSimply perform the calculation \n\n-log[concentration H+ or OH-]\n\nthis will give you a pH value if you used H+ ions, or a pOH value if you used OH-. To get the pH from a pOH, you must take the value and subtract it from 14. (14-pOH = pH)
950979
pH = - log[H+], so [H+] = 10^(-pH)\n\nfor pH = 7, [H+] = 10^(-7)
264737
Dissolve the following in 800ml distilled H2O.\n8g of NaCl\n0.2g of KCl\n1.44g of Na2HPO4\n0.24g of KH2PO4\nAdjust pH to 7.4 or 7.6 with HCl.\nAdjust volume to 1L with additional distilled H2O.
859846
That is all a matter of how you look at it. pH is the negative log of the hydrogen ion concentration. We say if something has a pH of less than 7, it is acidic. Soda contains carbonic acid and phosphoric acid, and thus has a pH less than 7. On the other hand, the sum of the pH and the pOH should equal 14. pOH is defined as the negative log of the hydroxide concentration. So, if the pH is 5, the pOH is 9, thus showing OH- to be present.\n\nPotato, potato....glass half empty, half full kinda thing...
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-]
907196
The concentration of H+ is 3.0 x 10^-4, so you know that the pH will be around 4.  What you need is a calculator that has a LOG button to convert the 0.0003 to an integer value.\n\nSo on your calculator you press the LOG button and then type in 0.0003 and then a ")" and press ENTER.  \n\nYou should get a result that says: -3.522878745 or so.  You need to round off to -3.5 and since pH is the NEGATIVE log of the hydrogen ion concentration you get rid of the minus.\n\nThe answer is 3.5.\n\n1x10^-3.5 is approximately equal to 0.0003.
844701
1) Calculate number of moles of NaOH:\n8.75 g/ (22.99+16.00+1.01) = 4.571 mol\n\n2) Calculate molarity in mol /liter\n4.751 mol / 25.4 mL = 0.187 mol/mL = 1.87 x 10^-4 mol/L
805565
Ca + 2H2O yields Ca(OH)2 + H2\n\nBasically it's almost balanced already, just add a 2 in front of the H2O.