Please estimate the pH 0f a 0.1 M solution of a diprotic acid with?
819244
a) Acid +H2O -> H3O(+)+ Basis\n\nKa1 =  x²/(0.1-x) = 10-4\nx<< 0.1 so x = 10-3\n\nso pH = 3\n\n2) second dissociation\nKa2 = y (10^-3 + y) / (10^-3 -y)\ny << 10¨-3\ny = Ka2 = 10^-8
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
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.
692123
1. pH = -log [H+] which is the hydronium ion concentration.  [H+] = 10^-2.85 = 0.00141 M\n[OH-] = 1 x 10^-14/0.00141 = 7.09 x 10^-12\n\n2.  pH = -log 12 = -1.08\n\n3.  neutral water would be when [H+] = [OH-], so [H+] would be the square root of Kw, so [H+] = 1.58 x 10^-7.  pH = 6.80.
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.
621533
NaOH ----------------> Na+   +   OH-\nFind the number of moles of NaOH\nMoles of NaOH = 0.10M * 0.150L = 0.015 moles\nSince the coefficients in fornt of NaOH, Na+ and OH- are all 1, therefore if 0.015 moles of NaOH disssociates, 0.015 moles of Na+ and 0.015 moles of OH- are produced.\n\nSimilarly we will work with HCl\nHCl --------------------> H+      +      Cl-\nMoles of HCL = 0.30M * 0.05L = 0.015 moles\nThis means the 0.015 moles of H+ will be produced\n\nSince H+ and OH- have equal amiunts, they will completely neutralize each other leaving the solution at neutral pH. Since the neutral pH is 7, therefore, the pH of this solution will be 7.
946866
11/27:Please post the answer so I can find the error in my work or determine if the answer key is wrong.  Lets see if we can get to the bottom of this one!\n\n\nFirst find the masses of each acid:\n\nHCl:  48.8 ml x 1.19 g/ml x 0.38 = 22.1 g\n\nHNO3:  17.5ml x 1.42 g/ml x 0.70 = 17.4 g\n\nSecond, convert grams to moles:\n\nMass HCl = 35.45 + 1.01 = 36.46 = 36.5 g/mol\nMass HNO3 = 16.0 x3 + 14.0 + 1.0 = 63.0 g/mol\n\nmoles HCl = 22.1 g/36.5 g/mol = 0.605 mol HCl\nmoles HNO3=17.4g/63.0 g/mol = 0.276 mol HNO3\n\nSince you're diluting to one liter, these are also the molarities.\n\n[HCl] = 0.605 mol/L ; [HNO3] = 0.276 mol/L\n\nEach of these acids are strong acids and can be assumed to dissociate completely.  If they were weak acids we would have to account for the acid dissociation equilibria here.  \n\nEach of these acids contribute only 1 equivalent of hydrogen ion.  Thus [H+] is the sum of the two concentrations.\n\n[H+] = [HCl] + [HNO3] = 0.605 + 0.276 = 0.881 mol/L = 0.88 M\n\nReport two sig figs since your mass % vals only have two sig figs.\n\nThe equilibrium equation for water, H+ and OH- is:\n\nH2O <--->  H+  + OH-   K = 1.00 x 10^-14\n\nH2O is a pure liquid and by definition has an activity of unity, so the equilibrium expression is:\n\n[H+][OH-] = 1.00 x 10^-14\n\nNow we can solve for [OH-] :\n\n0.881 X [OH-] = 1.00 x 10^-14\n\n[OH-] = 1.14 x 10^-14 mol/l = 1.1 x 10^-14 M\n\nAgain, two sig figs in final answer.\n\nFinally, pH = - log [H+]\n\nso pH = - log 0.881 = 0.055
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.
358982
NaOH is completely dissociated when it is dissolved in water.  A .0025M KOH solution therefore produces a [OH-] equal to .0025M.  The ion product of [H30+] and [OH-] is a constant, 1.0 x 10^-14 M^2.  By substitution, the [H30+] can be determined.  The pH can then be calculated.\n\n[H30+] = 1.0 x 10^-14 M^2/[OH-] = 1.0 x 10^-14 M^2/2.5 x 10^-3 \n\naaaand all that equals .4 x 10^-11... and you put that in correct scientific notation... 4 x 10^-12 M for the concentration of [H30+].\n\nWe then plug that in...   \n\npH= -log[H30+] = -log (4 x 10^-12) = 11.4\n\nThe pH of a .0025M solution of KOH is 11.4, which is greater than 7, making it a basic solution.\n\nStrong base at that.\n\nP.S.- If it is supposed to have the correct number of significant figures/digits, the answer would then be 11.
452992
Well first of all you have not given us all the information to be able to answer this buffer question.  We need to know the Ka (6.2x10-8)of the acid.  But the steps to perform the task.  Calculate how much of the acid is neutralized by the addition of the base.  Recalculate the equilibrium concentration of both the acid and base with the new acid value.  using Ka=[H+][Na2HPO4]/[NaH2PO4] solve for H+ and finally use pH=-log[H+].  Your final answer..pH=7.25
952215
[H3O+][OH-] = 1.0 x 10^-14 M always at 25C\nso if [H3O+]= 5.0 x 10^-10M\n[OH-]= 2.0x10^-5
976396
1 G, I dont see how it can be zero otherwise it would float off!\n\nOK call me an idiot...I have just read the other answers lol
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.
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).
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+]
411083
If the pH is 4, the pOH is 10 (pH + pOH = 14).  10^(-pOH) = [OH-].\n\nAnswer is 1E-10
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.