how much water to mix with sodium bicarbonate to neutralize battery acid.?
929306
When I do this, I use mostly sodium bicarbonate and very little water (just enough to dissolve the baking soda). The point is that the acid is very damaging, so you want to make darn sure that you've neutralized it. The baking soda is not dangerous, so you don't worry if you use excess baking soda. \n\nWhen you're done, and have neutralized the acid, you rinse everything thoroughly with plain water.
211023
wow, well put mrjeffy! I have one thing to add though. Because all acids and bases are electrolytes, you can use pH to determine what is and isn't an electrolyte in a solution. If it come out as acidic or basic, it is and electrolyte(the darker, the more basic or acidic). If it comes out neutral, like Distilled Water, it is not. Also don't forget that non-electrolytes will not completely dissolve into solution in water.
802509
Baking soda will neutralize the sulfuric acid.\n\nBatteries do not produce sulfuric acid!\nSulfuric acid is placed in the battery to generate eletricity through electrolysis.\n\nAs far as making the lead useable, they do recycle car and lead/acid batteries, is that useful enough?  \n\nIf you are planning on opening a car battery you are asking for potential injury or death.  You could cause an explosion, end up blind, maimed, or possibly dead.  I have seen batteries explode and it is not pretty.  I have had a few rupture/burst very close to my face while working on cars, that is not a nice situation to be in either.  Fortunately I wasn't hurt except for superficially, but batteries can be very dangerous.\n\nIf you need lead metal for a project or experiment, go ask a tire shop for their old, throwaway wheel weights.  They usually have tons of them and don't mind giving them away.\n\nDivers used to use lead weighted belts to keep them from floating, maybe there's some salty old sea man around a dock or something who knows where you can find some.
684804
use bicarbonate of soda,its short term though
511587
Baking Soda and Water!!
907209
Na2CO3 + NaHCO3  \n\nmoles Na2CO3 = 0.070L * 5.5 mol/L = 0.385 mol\nmoles of Na+ from Na2CO3 = 0.77 mol\nmoles NaHCO3 = 0.030L * 1.0 mol/L = 0.030 mol\nmoles of Na+ from NaHCO3 = 0.03 mol\ntotal moles of Na+ in the solution = 0.80 mol\nsodium ion (Na+) concentration = 0.80 mol/0.1L = 8.0M
744031
Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2\n\n0.1L * 0.2 mol/L = 0.02 mol HNO3\n\n0.02 mole HNO3 * 1 mol CO2/2 mol HNO3 = 0.01 mol CO2\n\nNow plug 0.01 mol CO2 into PV=nRT, using STP conditions.\n\nPV=nRT\n\nV = nRT/P\n\nV = (0.01)(.0821)(298)/(1)\n\nV = 0.24 L\n\n\n*** The person above me did it WRONG!  He confused sodium carbonate with sodium bicarbonate.  Also, he completely forgot to do the stoichiometric conversions.  He solved for moles of HNO3, and then plugged it into PV=nRT.  You must first convert moles of HNO3 to moles of CO2 with stoichiometric conversions, using the BALANCED equation.
773833
2NaOH + H2SO4  = Na2SO4 +2H2O                                      Sodium Hydroxide is 0.5moles.Two Moles neutralise One mole of Acid.   0.5Moles neutralise of Alkali neutralises 0.25Moles of Acid                                                                                           50mls of 1M SULPHURIC ACID = .05M THEREFORE MOLAR CONCENTRATION IS 5
706351
KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)
418463
Acid bath, zink and dc current
119509
my question is:\n\nNaHCO3(s) + NaOH(aq) -> Na2CO3(aq) + H2O(l)\n\n3.25 moles of Na2CO3 is produces.
586833
the vinegar is a solution of water and acetic acid.  the acetic acid exists as acetate ion and hydrogen ion.  the bicarbonate is a sodium ion and a bicarbonate ion.  when the two combine, the postive hydrogen ion combines with the negative bicarbonate ion to produce carbon dioxide and water, and the negative acetate ion and the postive sodium ion combine (well, they really don't combine, they stay apart and disolved in the solution) to become sodium acetate.  what's left is a water solution with sodium acetate desolved in it and bubbles of carbon dioxide coming out of the solution.
506202
Write out the balanced reaction which occurs.\n\nSince both NaOH and H2SO4 are strong electrolytes (strong base, strong acid), we can assume both dissassociate completely and the reaction produced Sodium Sulfate and water,\n\n2NaOH + H2SO4 --> 2H2O + Na2SO4\n\nFind limiting reactant in order to find which  will be left over (either NaOH or H2SO4).\nIn order to do this, you need to know both the concentrations of the NaOH and the H2SO4 in order to find the initial number of moles of each.\n\nOnce you know how many moles of either OH- or H+ are left over, you can then find the pH, or pOH like this,\n\npH = -log ([H+])\npOH = -log ([OH-])\n\npH + pOH = 14\n\n*Remember to take into account that the final volume of solution is the sum of the intial volumes of each othe reactant solutions when calculating pH.\n\nEDIT:\nYou know, it just occured to me, I dont know why I even brought up pH, that was not what the question was asking.  but everything I said still stands in order to calculate the molarity...you just dont have to go the one extra step and calculate the pH by taking the negative log of the molarity concentration of H+ ions.\n\nAlso, for all of you out there who are just given answers with not other explanation, it would be nice if you stated what the concentration was of....salt? NaOH?  H2SO4?\nBecause of this ommision, I would highly inclined to disreguard their answers as numerically incorrect as well.
885670
Get the chemical equation first:\nHCl + NaOH --> NaCl + H2O\n\nMoles of HCl used = (50/1000)dm3 X (1.0mol/dm3) = 0.05 mol\nMoles of NaOH used = (50/1000)dm3 x (1.0mol/dm3) = 0.05 mol\n\nFrom chemical equation, 1 mole of HCl needs 1 mole of NaOH for complete neutralization.\nFrom what you are given, you have the exact amount for complete neutralization (meaning, there is no excess reagents).\nFrom chemical equation again, 1 mole of HCl or NaOH produces 1 mole of water.\nThus you had exactly 0.05 mole of water produced here.\nHope this helps. :)
732031
Sodium bicarbonate in solid form is a poor conductor.    However, it will ionize when dissolved in water and conduct electricity then.
787481
NaHCO2 + H2SO4 --> Na2SO4 + CO2(g) + H2O(L)\n\nSodium bicarbonate + sulphuric acid equals sodium sulfate + carbon dioxide + water
84557
A simpler formula will be SO2+NO2+CO2+H20 (unlimited)-> H2SO4 (strong and oxidant) +  HNO3 (nitric acid, stronger and oxidant) +  H2C03 (carbonic acid, weak but ´till corrosive, just like the soft drinks)