how do i calculate [H] and [OH] CONCENTRATION USING THE Ph \nof wine which is 3.40
205398
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
353001
The H3O+ concentration has to be the same as the acetate ion concentration, that is 0.425. Since you don't tell the units I cannot calculate the pH. Why do you need to know pKa of acid if you already know the concentration of the acetate ions? This is a chemistry question. Why don't you place it under chemistry?
88853
For part a, I think you would need to know more information about the final buffer such as the final pH.\n\nFor part b, there are .020 mol of H+ ions in the solution because HCl is completely soluble. The molarity of the solution will be .020mol/1.00L = .020 M. To find pH, use -log(molarity). -log(.020) = 1.70
525814
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
466512
Here is what you need.\n\n2HF(aq) + Mg(s) ---> MgF2(s) + H2(g)
448010
Write the balanced chemical equation:\nNaOH + HNO3 -----> H2O + NaNO3\n\nthis means that 1 mole of NaOH is required to use up 1 mole of HNO3\nthis means that all of your HNO3 will be used up so to determine the excess NaOH all you have to do is subtract the # of moles of HNO3 from the number of moles of NaOH giving 0.00193 moles NaOH left.
251519
U ripped it of from da AISSE 2006 sample que paper didn't ya\na) pH = —[H+] ½\n= — log (1 x 10—10)\n= — (—10)\n= 10 ½\nb) basic ½\nc) decrease ½\nSince the concentration of H+, [H+] increases, — log10 [H+] decreases,\nso pH decreases
529743
[H3O+][OH-] = 1.0 x 10^-14 M always at 25C\nso if [H3O+]= 5.0 x 10^-10M\n[OH-]= 2.0x10^-5
481687
40 ml 3% H2O2 contains 3% H2O2 which is 1.2ml\n.\n20 ml 3% H2O2 means it has 20*3%= 0.6ml H2O2\nso 20ml 3%H2O2 + 20ml distilled water contains 0.6ml H2O2 in 40ml. So, 0.6/40=1.5%H2O2\n\nsimilarly, 10ml 3%H2O2 has 0.3ml H2O2.\nSo 10ml 3% H2O2 +30ml dist Water makes a 40ml solution with 0.3 ml H2O2 making it 0.3/40=0.75%H2O2 solution.\n\nSame steps for the 5ml case.
416783
We need x liters of 40% solution\nWe need (5-x) liters of 90% solution\n\nx(40/100) + (5-x)*(90/100) = 5 * (70/100)\n.40x + .90(5-x) = 5*.70\n.40x + 4.5 - .90x = 3.5\n.40x - .90x = 3.5 - 4.5\n-.5x = -1\nx = -1 / -.5\nx = 2\n2 gallons of 40% solution and 3 gallons of 90% solution.\n\nTo test our work:\n2 gallons of 40% solution gives you 2 * .40 gallons of alcohol = .8\n3 gallons of 90% solution gives you 3 * .90 gallons of alcohol = 2.7\n.8 + 2.7 = 3.5 total gallons of alcohol, with 1.5 gallons of water (I assume water since it's not mentioned what the other substance would be).\n\n3.5 / 5.0 = .70 or 70%
480976
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])
443139
Assuming KOH is a strong base,\npOH = - log [-OH] = - log (1.00 M) = 0\npH + pOH = 14\npH = 14 - pOH\npH = 14\npH = - log [H+]\n[H+] = antilog(-pH)\n[H+] = 1 x 10^-14 M
402503
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
247376
I don't know if this is right ... but check it and tell me\n\nV*C KOH = V*C HBr\n\nV KOH = 18.6ml C KOH = 0.752M\nV HBr = 22.1ml C HBr = ?\n\n18.6*0.752 = 22.1*C HBr\n\nC HBr = 0.6329M\nIts seems reasonable cause the difference between concentrations isn't significant
135221
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+]
402793
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
441852
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.