Calculate the [OH-] of each aqueous solution with the following [H30+]?
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
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.
48368
2HC2H3O2 + Ba(OH)2 --> Ba(C2H3O2)2 + 2H2O
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.
844168
This is very simple\n\nCa+H2O -> Ca(OH)2 + H2\n\nThe reactant has\n1 atom of Ca\n1 atom of H\n2 atoms of O\n\nThe product has\n1 atom of Ca, ok\n2 atoms of H\n2 atoms of O\n\nJust put a coefficient of 2 in the water (H2O)\n\nSo the chem. eqn becomes\n\nCa+2 H2O -> Ca(OH)2 + H2\n\nThat's it.\nThe equation is already balance
639844
[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
674948
HF + H2O => F + H3O+\nNH4+ + H2O => NH3 + H3O+\nH2CO3 + H2O => HCO3- + H3O+\nHNO2 + H2O => NO2 + H3O+\n\nbasically, the oxonium ion is formed in each one, when a proton is donated to water.
103867
NaClO + 2FeSO4 + H2O -> NaCl + 2 Fe(3+) + 2 (SO4)(2-) + 2 OH(1-)
904934
2HNO3 + Mg(OH)2 -->Mg(NO3)2 +2H20\n\n2H+(aq) + 2NO3-(aq) + Mg+(aq) + 2(OH)-(aq) -->Mg+(aq) + 2(NO3)-(aq) +2H20(l)\n\n2+(aq) + 2(OH)-(aq) --> 2H20(l)
752218
2KOH + Cl2 ----> KCl + KClO + H2O\n6KOH + 3Cl2 ----> 5KCl + KClO3 + 3H2O
470703
reduced: 2H2O + 2e- ---> H2 + 2OH-\n\noxidized: 2I- ---> I2 +2e-
923888
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
224213
well, pH = -log[H+]. so if you just undo the log you will have the concentration of H+ (which is the same as H30+). then .. [H+} + [OH-] = 14 good luck!
959805
HBr, -1\nHOBr, +1\nBr2, 0\nHBrO4,  +7\nBrF3  +3\n\nThere is nothing to explain.  There are rules, and you need to learn them and follow them.\n\nHere is a site with a good set of rules\nhttp://www.science.uwaterloo.ca/~cchieh/cact/c123/oxidstat.html
704589
Hydronium ions, in an aqueous solution.  H30+
813911
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
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.