what would be the pH of a solution when [H3PO4]=[H2PO4-]?
117935
I believe that it is 7... If the concentration of the acid and base are equal then they would probably neutralize the solution.
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
766811
d. \n\nIt's a double replacement (acid/base) reaction.  The Sr and the H are both positive and they switch.  Since Sr has a 2+ charge and the PO4 has a 3- charge, you use a common multiple of 6.  The other product would be H(OH) or water.
908950
we have this equation :\n2PO4 3- + 3Pb2+ = Pb3(PO4) 2-\n\nthe moles of Pb2+ = the moles of Pb(NO3)2 = 0.160 * 0.150 = 0.024\nthe moles of PO4 3- = the moles of Na3PO4 0.024 * 2 / 3 = 0.016\nthe Volume of Na3PO4 = 0.016 / 0.2 = 0.08 L = 80 mL
438792
Okay. This is a multi-step pH problem. First, find the Kb of the fluoride ion (F-). To do this, follow this formula:\n\nKw/Ka = Kb\n\nThis formula can be rearranged to be Kw/Kb = Ka, by the way. Kw is the K of dissociation of water, which is 1.0e-14.\n\nPlug in the numbers and you get 1.47e-11 as a Kb for fluoride. Now, write an equilibrium ICE table, filling in the values you know.\n\n_F- + H2O --> HF + OH-\nI\nC\nE\n\nSolve the equilbrium table and you get the following equation:\n\nx^2/(0.00339-x) = 1.47e-11\n\nx is the concentration of both HF and OH-.\n\nSolve the equation and you get 2.23e-7 as the concentration of OH-. The easiest way to finish the problem now is to find the pOH. Take the -log of the concentration of OH-. The pOH is 6.651. Now, follow this equation: pH + pOH = 14. The pH of the solution is 7.34876. To the right amount of significant figures, the answer is 7.349 (the numbers to the left of the decimal place doesn't count toward the sig figs in pH's).\n\nHope this helped.
47432
3AgC2H3O2 + Na3PO4 -----> Ag3PO4 + 3NaC2H3O2
929172
3Mg(NO3)2 + 2Na3PO4 => Mg3(PO4)2 (prec.) + 6NaNO3\n\nAqueous solution is assumed.  The magnesium phosphate is insoluble and precipitates out.  This is normally indicated with a downward arrow right after its formula, but no such symbol exists on this keyboard.
812798
3Ca(OH)2+2H3PO4>Ca3(PO4)2+6H2O
834810
You have 75mL of 0,4000M NaOH solution\n0,4000 M= moles of NaOH/0,075L\nmoles of NaOH = 0,030\n 3NaOH + H3PO4 → 3H2O + Na3PO4\nNaOH reacts with H3PO4 at a ratio of 3 moles : 1 mol\nTherefore: 0,030 m NaOH*(1m H3PO4/ 3m NaOH)= 0,010 moles of H3PO4\nThen\nM = 0,010 moles/0,050L\nM = 0,20\nYou are correct; nevertheless, watch the figures!!!!
696739
The chemicals are not correct\n\n2 (NH4)3PO4+3 Pb(NO3)2-> Pb3(PO4)2+6 NH4NO3?
494638
P3O10 does not exist; it's P4O10.\n\nBalanced equation is:\n\nP4O10 + 6H2O --->  4H3PO4
541583
pH from 0 to 7: Acid solution.\nPH from 7 to 14: Basic solution.
895751
Al2S3 + 6H2O -> 2Al(OH)3 + 3H2S\n\nP4O10 + 6H2O -> 4H3PO4\n\n\nShould be right.
906651
2KOH=>K(2)O+H(2)O\n\nFeO+C=>CO+Fe\n\nCa+2AlCl=>CaCl(2)+2Al\n\nP(2)O(5)+3H(2)O=>2H(3)PO(4)\n\n2KI+Pb(NO3)2=>2KNO(3)+PbI(2)
325149
3NH3(aq) + 3H20(l)--> 3NH4+(aq) + 3OH-(aq)\n AlPO4(s) --> Al3+(aq) + PO4 3-(aq)\n\nequal:\n\nAlPO4(s) +3NH3(aq)+3H2O(l)-->Al(OH)3(s) + (NH4)3PO4(aq)\n\nliquid fertilizer is  (NH4)3PO4(aq)
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.
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.