Please help chemistry question pH levels??
960642
Look up the chemical in a pKa ( -log of dissociation constant K)table of acids. H2PO4- has a pKa of 7.20. It is not very acidic.
646163
Presuming that the base is monobasic.\n\na) with 25 ml added, you are at the equivalence point.  The pH is 7.\nb) after adding 25.01 ml, you will have 0.01 ml of base that did not get reacted.  The molarity of the base in solution will then be 0.1 * (0.01/50.01) = 2.0 x 10^-5.  Your pOH is 4.70, pH would be 9.30\nc) Now, it's 0.1 ml of unreacted base.  Molarity is 0.1 * (0.1/50.1) = 2.0 x 10^-4.  Your pOH is 3.70, pH is 10.30.\nd) Now 1 ml of unreacted base.  Molarity is 0.1 * (1/51) = 2.0 x 10^-3.  pOH is 2.70, pH is 11.30.\ne) Now 25 ml of unreacted base.  Molarity is 0.1 * (25/75) = 3.33 x 10^-2.  pOH is 1.48, pH is 12.52.\n\nIn each case, you have to figure out the amount of unreacted titrant and recalculate the molarity as if you were simply diluting it, using McVc = MdVd, where the volume is now the total volume of the solution (the unreacted amount plus the 25 ml of acid and the 25 ml of base that have neutralized each other).
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.
133368
Normal ph level is 7.35-7.45
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-]
1095569
check this link-- http://www.orgsyn.org/orgsyn/prep.asp?prep=cv6p0468.\n\ninstead of Me use Ph
768732
pH at pKa +- 1 unit. from henderson hasselbalch\npH = pKa + log [base]/[acid] \ngiving +- 1 when the ratios are 1/10 or 10/1 in concentration\nthe capacity of a buffer is achieved when the ratio of weak acid to conjugate base exceeds 10:1 or vice versa. \nDoes that help?
832121
Nope, :s sorry\n\nIts the same as before. The K(w) of water is 1.0 *10^-14\nK(w) = [H+][OH-]\n1.0*10^-14=[H+][2.8*10^-4]\n[H+]= 3.57*10^-11\n\npH=-log[H+]\n the pH is 10.4\n\nand pOH =-log[OH-] u have that in the ques\n\nHappy trails
133652
pH 6.5 sounds about right - it depends where you live and the degree of chlorination, etc.\n\nTry measuring it.
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...
625417
pH= -log(6.0 x10^-3)\npH is about 2.22
378003
pH for a salt of strong base and weak acid is given by:\n         pH = 7 - (1/2)pKb - (1/2)log C\nwhere pKb = dissociation const of strong base\n       C  = concentration of Acid.\n\nKb = Kw / Ka = (10^-14) / (6.5 x 10^-5) = 1.53x10^-10\npKb = -log (1.53x10^-10) = 9.8154\nlog C = log 10^-1 = -1\n\npH = 7 - (1/2)x 9.8154 - (1/2)x-1\n   = 7 - 4.9077 + 0.5\n   = 2.5923
703155
No surprise at all, the transition does not take place at an pH of exactly 7 but already before. Take a look here for a graph:\n\nhttp://antoine.frostburg.edu/chem/senese/101/acidbase/faq/indicating-strong-or-weak.shtml
664234
NH4ClO4 --> NH4^+ + ClO4^-\n\nNow in a water solution,\nClO4^- + H2O --> HClO4 + OH- --> This doesn't happen because HClO4 is a strong acid.  Therefore the only contribution to the pH is from the NH4^+\n\nNH4^+ <=> NH3 + H^+\nYou start with [NH4^+] = 0.40 M, and [NH3]=[H^+] = 0 M; after undergoing equilibrium [NH4^+] = 0.40 - x, while [NH3]=[H^+] = x; Ka = [NH3][H^+]/[NH4^+] = (x)(x)/(0.40-x) = Kw/Kb = (1 x 10^-14)/(1.8 x 10^5)\nAssume x << 0.40, so x^2/0.40 = 5.56 x 10^-10\nx = 1.49 x 10^-5 M = [H^+]\npH = -log(1.40 x 10^-5) = 4.827
814606
pH samples need to be analyzed immeditely for regulatory purposes.  There is no holding time.  They must be test ASAP.
1118024
Acidic \n\nbelow 6 acidic above 7 basic
887006
pH= -log([H30+])\n\n\njust put the #'s in the above eqn.\n\nfor pOH, its pOH= -log([OH])\nand for pH given [OH], its pH= 14 - (-log([OH])