what would the pH and pOH if [H]=6.8x10^-4?
597779
The first answer is poh=3.167 and the ph=10.832\nThe second answer is poh=2.119 and the poh=11.881\n\nNot sure onthe concentration of hydrogen though.
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
1108880
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
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.
1121634
The pH of NaOH will depend on the concentration of NaOH.\nWithout the concentration of NaOH, one can only state that it has pH of greater than 7, since NaOH is basic.  \n\nNaOH is considered to be completely soluble so the concentration of OH- will equal to the concentration of NaOH in Molarity.  \n\nOnce [OH-] is determined, you can use the equation:\npOH = -log[OH-]  \n\nOnce pOH is determined, use following equation:\n14 - pOH = pH\n\nThat will give you the pH of the NaOH at given concentration.
970544
Here is what you do: \nDo the same operations as I did on the equations with the H values and you will be OK (Do not forget to use the same multipliers on the H as I did on the reactions).\n\n10PCl3 + 5O2 ---> 10Cl3PO\n10PCl2 + 10Cl2 --- 10PCl5\nSubstract the equations to get (keep coeff positives by moving to the other side):\n\n5O2 + 10PCl5 ---> 10Cl3PO + 10Cl2\nP4 + 5O2 ---> P4O10\nSubstract and get\n10PCl5 + P4O10 ---> P4 + 10Cl3PO + 10Cl2\nP4 + 6Cl2 ---> 4PCl3\n\nSum the two to get (P4 goes):\n\n10PCl5 + P4O10 ---> 10Cl3PO + 4Cl2 + 4PCl3\n\n4 PCl3 + 4Cl2 ---> 4PCl5\n\nSum the two equations\n\n10PCl5 + 6P4O10 ---> 10 Cl3PO.
473164
You have to make sure all the elements on each side of the equation are equivalent. \n\n4 PH3 + 8 O2 = P4O10 + 6 H2O\n\n\nThus each side of the equation has total:\n4 Phosphorus\n16 Oxygen\n12 Hydrogen
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.
541583
pH from 0 to 7: Acid solution.\nPH from 7 to 14: Basic solution.
511319
NaOH is in excess\n\nSubtract moles of H+ from OH- (.0345 mol OH- - .025 mol H+ = .0095 mol OH-)\n\nDivide by total volume (.0095 mol OH-/.0595 L = 0.1596 M)\n\nFind pOH = -log [OH-] = -log [.1596] = 0.7967\n\nFind pH = 14-.7967 = 13.20
657714
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...
859846
That is all a matter of how you look at it. pH is the negative log of the hydrogen ion concentration. We say if something has a pH of less than 7, it is acidic. Soda contains carbonic acid and phosphoric acid, and thus has a pH less than 7. On the other hand, the sum of the pH and the pOH should equal 14. pOH is defined as the negative log of the hydroxide concentration. So, if the pH is 5, the pOH is 9, thus showing OH- to be present.\n\nPotato, potato....glass half empty, half full kinda thing...
650541
pH = -log ([H3O+])\npOH = -log ([OH-])\n\npH + pOH = pKw = 14\npOH = 14 - pH\n\n[OH-] = 10^-(pOH)\n\nPlug in and solve.
264737
Dissolve the following in 800ml distilled H2O.\n8g of NaCl\n0.2g of KCl\n1.44g of Na2HPO4\n0.24g of KH2PO4\nAdjust pH to 7.4 or 7.6 with HCl.\nAdjust volume to 1L with additional distilled H2O.
901261
I think that the equation is \n3Ba(OH)2 + 2H3PO4 ---> Ba3(PO4)2 + 6H2O\nbut maybe the answer is\nBa(OH)2 + 2H3PO4 ---> Ba(H2PO4)2 + 2H2O\nbut no problem you can use the same  way to calculate.\n\n3Ba(OH)2 + 2H3PO4 ---> Ba3(PO4)2 + 6H2O\nthe moles of H3PO4 0.01015 * 0.141 = 1.43 * 10^-3\nthe moles of Ba(OH)2 needed : 1.43 * 10^-3 *3 / 2 = 2.145 * 10^-3\nthe volume of Ba(OH)2 needed : 2.145 * 10^-3 / 0.0584 = 0..03673 L = 36.73 ml
781203
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
796693
4PH3 (g)  ---> P4(g) + 6H2(g)\n\n4 moles of PH3 are consumed then 1 mole of P4 is produced and 6 moles of H2 are produced.\n\nIf you are consuming 0.0048 mol/s of PH3, then by the relation above you will be producing:\n\n0.0048/4 = 0.0012 mol/s P4\n0.0048*6/4 = 0.072 mol/s H2.