THe problem I have is that the regular pH equations that I know, don't apply!\nI did this experiment and  I found the solution to be acidic, but theoritically none of the 2 ions undergo hydrolosis, so the solution should be at pH7, right? I have been told that the PH was not as I expected because of the high charge density of the Fe3+ ion, but is there a formula to calculate a pH for this case?
122218
FeCl3 is a Lewis acid, because the Iron loses electron density because of the chlorines. So it complexes with the lone pair in water on the oxygen atom. In doing so, a proton is released from the water, which then complexes with the other water molecules to make H30+
195370
Uh, the rest of the question isn't appearing. But it's an equilibrium solubility problem. Write a reaction:\n\nMgF2 --> Mg+2 + 2F-\n\nKsp = 5.52e-9\n\nFill in the values on the ICE table (I don't know them, the numbers are missing), and set an equation up.\n\nKsp = [F-]^2 * [Mg+2]\n\nSolve with the numbers from the ICE table. Remember that those concentrations in the above equation are the equilibrated concentrations, not the initial.\n\n\nHope this helped.
629476
HCl   +  NaOH   ===   NaCl   +   H2O\n0.0015 ,  0.0016 ,             0        ,       0        at start of reaction (moles)\n0,           0.0001 ,            0.0015 ,    0.0015  at end of reaction(moles)\n\nevery mole of water contributes about 10^-7(ten to the power -7)\nmoles of OH-  and H+ . thus their contribution to the pH is neglected when there is a considerable amount of acid or base present  i.e more than 10^-5 moles/litre (not moles but CONCENTRATION)\nThus concentration of NaOH = 0.0001/90*1000 moles/litre{total volume = 50+40 ml} = 0.001 .\nas NaOH is a strong electrolyte like any other sodium compound  we can assume complete ionisation . thus[OH-]=[NaOH] = 0.001. now pH = -log[H+]=14-(-log[OH-])= 14+log[OH-]  =14+log[0.001]=11\nthus the pH=11
318012
NaOH + HCl --> NaCl + H20\n\nn(NaOH) = 1.00 x 32.6cm3\n        = 1.00 x 32.6 x 10^-6\n        = 3.26 x 10^-5  mol\n\nn(HCl) = 1.00 x 25cm3\n       = 1.00 x 25 x 10^-6\n       = 2.5 x 10^-5  mol\n\ntherefore, NaOH is in excess.. hence the final solution is alkaline, which means the pH is more than 7.\n\nDue to the limitation of my knowledge, i can't give you the exact pH value. Just hope that it will help you to continue solving...
598186
I think this is how you do it:\nacid and base is measured in terms of pH and pOH. as you know, the H makes it acidic and the OH makes it basic, and if you mix an H and OH, you get water, which is to say it was neutralized.\nSo, let's say you have HBr with a pH of 4. You that pH is a measure of the concentration of H present in the mixture. So, to get the actual number you do -10ph, and you'll get a number like 10^-5 or something. So, to neutralize that many Hs, that's how many OH's you need. so, you'll need a pOH 4, which is the same thing is a pH of 10 (pOH = 14 - pH). \nhope that helps, good luck. and read the book, and do some of hte exercised in the book.
351460
I am not a chemistry student but I make this research for you and maybe it helps you http://www.chem.latech.edu/~deddy/chem104/104Antacid.htm\nhttp://www.chemcool.com/regents/acidsbasessalts/aim3.htm\n\njust copy those links and you should be on your way to solve ur problem.
28552
Wow, I hope these are right.  I wish i had a solubility chart or something (Which you probably have)  \n\nCuSO4(aq) + Na2S(aq) ---> CuS (s) + NaSO4 (aq)\nNet ionic - Cu2+ (aq) + S2- (aq)  ---> CuS (s)\n\nNiCl2 + 2KNO3 --> Ni(NO3)2 (aq) + 2KCl (aq)\nnet ionic - none (if NiCl2 isn't aq, then this is wrong)\n\nCa(OH)2 + 2HC2H3O2 ---> Ca(C2H3O2)2 (aq) + 2H2O (l) \nnet ionic - H+ (aq) + OH-(aq) ---> H2O (l)\nactually, since the acetic acid is a weak acid, this might be a little different.  If you learned it this way, then your net ionic is:\nHC2H3O2 (aq) + OH- (aq) ---> H2O (l) + C2H3O2- (aq)\n\nHope this helped
342054
You are not giving us all of the info.\nYes the total or initial concentration of the acid is C=(8.345*10^-3)/0.05 = 0.1669 M\n\nFor a monoprotic acid Ka=[H+][A-]/{HA]\n[A-]=[H+]=10^-pH=10^-4\n[HA]=C-[H+]= 0.1669-0.0001= 0.1668\nso you have calcualted Ka correctly.\n\nKa depends on the nature of the acid and temperature (for small molecules whereas for bigger molecules like proteins it is also affected by the nature of the surrounding chemical groups).\n \nThus addition of base doesn't affect Ka, only the pH.\nIs the base monoprotic or not? If not, what's the valency? Is it strong or weak? If it's weak what's the Kb? Is it the conjugate base of the weak acid? Without these info we can't calculate the pH.
307901
[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
349315
This deals with Bronsted-Lowry (B-L) acids and bases, and conjugate-base pairs \n\nB-L acids: donate H+ ions: STRONG ACIDS dissociate into ions\nH20 + HCl --> Cl- + H+  (H+ is the same as hydronium [H3O+])\nso it can be written as H20 + HCl --> Cl- + H3O+ \nThis looks like it donates a H+ ion, but it dissociates because it is a STRONG acid.  \n\nwk acid: HA= wk acid; A- = conjugate base\nHA + H2O ---> H3O+ + A-\n\nB-L bases: accept H+ ions: STRONG bases dissociate\nNaOH + H2O --> H2O + Na+ + OH-\n\nwk base: B = wk base; HB = conjugate acid\nB- + H2O --> HB + OH-\n\nThat's all you need to know to write all eqns.
99119
Check out the buffer recipe calculator at <http://www.bioinformatics.org/JaMBW//5/4/index.html>\n\nSome things to consider:\n\nAt what temperature do you want the pH to be 7.5? (the pKa of tris-base varies as a function of T)\n\nWhat ionic strength do you need?
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])
558258
pH = -lg [H+]\n\nHence, \npH = 3 means [H+] = 10^ -3\npH = 5 means [H+] = 10^ -5\n\npH = 3 has 100 times more H+ ions in the same volume of solution than that of a pH = 5, which means it is 100 times more acidic
510285
The strong base NaOH will be fully dissociated in water, and you're adding quite a bit (looks like 5 grams/10 mLs).  The water actually consists mostly of hydroxide anions at this point, and hypochlorite ions (your oxidiser) from the bleach.  The pH is 14 or so.  These are the only conditions you'll find Fe(OH)3 at.\n\nIt Ferric hydoxide is  instable due to its loose bonding (easy dissociation) with the hydroxide ions.
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
401787
HCL and NaOH are 'strong' acid and base. When they react, the product is neutral. NaOH + HCL --> NaCl + H2O\n\nWhat you need to find is which is the limiting reactant, the other one (what is left over) will effect the pH by its concentration in the otherwise neutral solution.\n\nFor your first problem:\n\n0.0557M HCL = 0.0557mols/liter*0.00542L = 3.02*10^(-4)mol HCL\n\n0.0377M NaOH = 0.0377mols/liter*0.00337L = 1.27*10^(-4)mol NaOH\n\n3.02*10^(-4)mol HCL - 1.27*10^(-4)mol NaOH = 1.75*10^(-4)mols HCL\n\n0.00542L + 0.00337L = .00879L\n\n[(1.75*10^(-4)mols HCL) * (1 L) ] / (.00879L) = .020M HCL\n\npH = -log(.02) = 1.70
153793
There are three species in attendance in this solution, H+ and ClO- ions (that come from the disassociation of hypochlorous acid) and hypochlorous acid itself. \n\nThe Ka value comes from the following equation: \nKa= [H+]*[ClO-]/[HClO]\n(the brackets stand for concentration)\n\nWe can set [H+] to the variable x, and because for each mole of H+ formed, one mole of ClO- is formed, [ClO-] is also x.  The concentration of HClO is initially .10M, and because x amount of HClO splits up into the two ions H+ and ClO-, the final concentration of HClO that we will be using is .10M-x.  \n\nThat gives us this following equation:  \nKa=x*x/(.10M-x) \nor \nKa=x^2/(.10M-x)\nand because Ka=3.5x10^-4, \n3.5x10^-4=x^2/(.10M-x)\n\nNow you can algebraically solve for the value of x, or use a graphing utility to do it for you.  \nCross-multiply:\n3.5x10^-4(.10M-x)=x^2\nNow distribute:\n3.5x10^-5 - 3.5x10^-4x = x^2\nand set it equal to zero:\n0=x^2 + 3.5x10^-4x - 3.5x10^-5\nand use the quadratic equation to solve for x:\nx= -3.5x10^-4 + sqrt((3.5x10^-4)^2 - (4*1*-3.5x10^-5)) all divided by 2*1\nx=.0057 \n\nSo the [H+] = [ClO-] = x = .0057 M.  \nThe HClO concentration is .10M-x=.094M.