How can a find the molar solubility of iron hydroxide at a ph of 12?
407734
Hummm:  You can calculate the [OH-] from the pH.  [OH-] = 10^-2.  pH + pOH = 14 (from [H+]x[OH-] = 10^(-14):  pOH = 14 - 12 = 2.  Find the solubility product for iron (III) hydroxide in a table in the back of your textbook or a handbook of Chemistry or physics. Kp = [Fe+3] x [OH-]^3. Substitute numbers in and solve for [Fe+3].  This is also the molar solubility as one mole of iron hydroxide contains one mole of Fe+3.
854123
I assume this is a chemistry question. That being the case, 1 mol is equal to 6.02 x10^23. Such as in one gram of iron, there are approximately 56(6.02 x10^23) atoms of iron. 56 is the molar number for iron. \n1 mol = 6.02 x10^23\n\nSo then, you're asking what is mol/dm^3.\ndm^3 is the volume of a litre. \nSo mol/dm^3 is mol/L or mols per liter\n\nIt is a simple concentration. How many mols of a substance is contained in one liter. \nThis value differs from one element to the next because the elements vary in size. \n\nSo mol/dm^3 is just the MOLARITY of a concentration.
103867
NaClO + 2FeSO4 + H2O -> NaCl + 2 Fe(3+) + 2 (SO4)(2-) + 2 OH(1-)
459258
start with computation of molarity of solution:\nM_BaOH = (51.24g/ MW BAOH)/ 1.2 L\n\nthen \nusing M1V1 = M2V2\nM_BaOH (x) = (0.1) (1L)\nsolve for x
734376
Since the equation you listed isn't balanced, I'm going to attempt that first:\n\nFe2O3 + 3(CO) ----> 2Fe + 3(CO2)\n\nsince 1 mole of Fe2O3 seems to produce 2 moles of Fe, it would appear we need 5.9/2 = 2.95 moles of Fe2O3\n\nGiven that 1 mole of these elements weighs: (see source)\nO = 16g\nFe = 56g\n\n2.95 moles * ( 2 * 56g + 3 * 16g ) = 2.95 * 160g = 472g\n\nHope that helps.
822325
a) Find some Acetic Acid (CH3COOH)\nb) Find some Copper (II) Hydroxide (Cu(OH)2)\nc) Add to form Cu(CH3CO2)2 + 2H2O\n\nWithout the volume or the number of moles, preparation of a 0.530M can be done several ways. \n\n2CH3COOH + Cu(OH)2 --> Cu(CH3CO2)2 + 2H2O\n\nUsing 1L:\n0.530 mol of Cu(CH3CO2)2 means 1.060 mol of CH3COOH\n\n1.060 / 0.5L = 2.120M\n\n0.530 mol of Cu(CH3CO2)2 means 0.530 mol of Cu(OH)2\n\n0.530 / 0.5L = 1.060M\n\nSo mix 500mL of 2.120M Acetic Acid with 500mL of 1.060M Copper (II) Hydroxide.\n\nN.B. Being a weak acid, acetic acid will not fully react with copper hydroxide, and hence the actual molarity requiredmay be higher.
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.
808549
M=molesof solute/Liter of solution\n\nM= molarity
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+]
727004
(NH4)2Cr2O7 -> Cr2O3 + 4*(H2O) + N2\n\n(NH4)2SO4 + 2*(NaOH) -> 2*(NH3) + 2*(H2O) + Na2SO4\n\nFe2(SO4)3 + 6*(NH3) + 6*(H2O) -> 2*(Fe(OH)3) + 3*((NH4)2SO4)
952998
The reductive dissolution of goethite ( -FeOOH) and hematite ( -Fe 2 O 3 ) by phenolic reductants has been examined in order to improve the understanding of iron transformations in soils, sediments and aquifers. Rates of goethite reductive dissolution by hydroquinone increased as the pH was increased from pH 1.8 to 4.65, arid the following reaction stoichiometry was obeyed: 2 - FeOOH + QH 2 = 2 Fe 2+ + Q + 4 OH - . As the pH was increased from pH 4.5 to 6.0, the reductive dissolution rate decreased to below the detection limit. At pH 3.4, the reductive dissolution of hematite was two orders of magnitude slower than goethite. The relationship between structure and reactivity was examined for a series of mono-, di-, and tri-hydroxybenzene reductants. Rates of reductive dissolution decreased in the following order: catechol ~- hydroquinone 〉 3,4-dihydroxybenzoic acid 〉 resorcinol-phenol-4-hydroxybenzoic acid.
356281
MgCO3 + 2HCl ---> MgCl2 + CO2 + H2O\n\n2HCl + MgO ---> H2O + MgCl2\n\nBarium hydroxide is a base so can be neutralised with any acid.  An example being HCl.\n\nBa(OH)2 + 2HCl ---> BaCl2 + 2H2O\n\nor H2SO4\n\nBa(OH)2 + H2SO4 ---> BaSO4 + 2H2O
633715
5. KHCrO4 ----------------> K+   +   HCrO4-\nSince KHCrO4 dissociates complteley,\n the molarity of HCrO4- = molarity of KHCrO4\n the molarity of HCrO4- = 0.05M\n\nNow HCrO4 ------------------>    H+    +     (CrO4)2-\nCalculate the concentration of H+ from the pH value\n[H+] = 10^(-pH)   = 10^(-3.80)  = 1.58 * 10^(-4)\nIt is clear the only source of H+ is HCrO4-, therefore the concentration of (CrO4)2- is the same as concentration of H+\n[(CrO4)2-]   =   [H+]   = 1.58 * 10^-4\n\nKa = {H+} * [(CrO4)2-] / [HCrO4-]\nKa = [1.58 * 10^(-4)]^2/0.05\nKa = 5.02 * 10^ (-7)\n\n6.       KOH----------------> K+    +    OH-\n         HClO4 -------------> H+    +    ClO4-\n(a) \n    Moles of KOH = Molarity * Volume = 0.200M * 0.030L\n    Moles of KOH = 0.006 moles\n    From the equation above, \n    Moles of OH- = moles of KOH = 0.006 moles\n    \n    Moles of HClO4 = Molarity * Volume = 0.150M * 0.030L\n    Moles of HClO4 = 0.0045 moles\n    From the equation above,\n    Moles of H+ = moles of HClO4  = 0.0045 moles\n\n     H+    +    OH-    ------------>   H2O\nWhen this reaction is finished, 0.0015 moles of OH- are left\n[OH-]  =  moles of remaining OH- / Total Volume\n[OH-]  =  0.0015 moles/ 0.06L (30ml + 30ml)\n[OH-]  =  0.025M\npOH = -log [OH-]\npOH = -log (0.025) = 1.60\npH = 14 - pOH  = 14 - 1.60 = 12.40 \n\nDo the same for parts b, c, d, and e\n\n7. I will assume that \n   H2S --------------------------> 2H+    +    S2-\n   Ksp of HgS = 2.0 * 10^(-53)\n   Ksp of FeS = 6.3 * 10^(-18)\n\nKsp of HgS = [Hg2+] * [S2-]\n[Hg2+] = 0.01 M\n[S2-] = x\n2.0 * 10^(-53) = x*0.01\nx = 2.0 * 10^(-51)\nFrom the above equation, for every 1 mole of S2- there are 2 moles of H+\ntherefore,\n[H+] = 2 * 2.0*10^(-51) = 4*10^(-51)\npH = -log (4*10^(-51)) = 50.39\n\nKsp of FeS = [Fe2+] * [S2-]\n[Fe2+] = 0.01M\n[S2-] = x\n6.3* 10^(-18) = x* 0.01\nx = 6.3 * 10^(-18)\nFrom the above equation, for every 1 mole of S2- there are 2 moles of H+\ntherefore,\n[H+] = 2 * 6.3 * 10^(-18) = 1.26*10^(-17)\npH = -log (1.26*10^(-17)) = 16.90\n\nthe pH range is from 16.90-50.39\nThis is the best explanation I can gibe. By the way, pH level beyond 0 and 14 do exist, but, I'm not 100% about this answer.\nI fixed it, I think it is correct now.
884150
Well, in pure water, if you let x = [Mg2+] the [OH-] = 2x.  Per the equilibrium expression 5.61 x 10E-11 = 4xE3 so x=2.4 X 10E-4 M\n\nIn the NaOH solution, the [OH-] is 0.180 M (the contribution from the Mg(OH)2 is negligible).  So in this case\n                5.61 x 10E-11 = [Mg](.180)E2  \nSolving for [Mg], I get 1.7 X 10E-9 M\n\nCheck my arithmetic, I often get careless punching the calculator.
844624
FeCl3   +    3NaOH  ----->    Fe(OH)3   +   3NaCl\n\nthats it
566520
Fe(s)+H2SO4(aq)excess--> Fe2+(aq) + SO42-(aq)+H2(g)\n\nratio is 1:1\n\n5gFe/55.85g/molFe= .090mol\n\nand\n\n1 molar solution =1 mole per liter\n\nso,\n\n.09mol(1 liter/1mol)= .09L or 90mls
157730
I think FePO4 +O2 >>>>> Fe2O3 + PO2. balance it and solve it . n its simple stoichiometry.