What volume of 3 M Sodium Hydroxide (NaOH) does it take to completely neutralize 700 mL of 6 M HCl?
478026
1400 mL of NaOH needed...\n\nneed to make sure the number of moles are the same in order for them to completely neutralize\n\nM = Molarity... number of moles of solute in one liter of solution\n\nif you have 3 M for the sodium hydroxide you would need twice that amount in order to neutralize the 6 M of hydrochloric acid\n\nyou would need 1400 mL since the number of moles of the sodium hydroxide is exactly half of that of the hydrochloric acid. With knowing that the number of moles NaOH is half, you would need double of what you have for the HCL... 700x2 is 1400 mL\n\nthanks redsuns :D
1136961
Are you preparing a buffer soln with NaHCO3 and NaCl ? Then take equimolar amounts of both and dissolve them in 100 ml of water and then dilute them to the desired strength.
840903
First find out how many  moles of NaOH in 5.18 g sample.\nmolar mass for NaOH is 23 + 16 + 1 = 40 g/mol\n\n5.18 g NaOH / 40 g/mol = 0.1295 mols NaOH\n\nNext calculate the molarity of the first solution. Molarity = mol/Volume (in L)   300 mL = 0.3 L\n0.1295/0.3 L = 0.4317 M\n\nFinally that solution is being diluted so find out how many mols you are taking out when you remove 25 mL uisng the same equation Molarity = mol/vol    26 ml = 0.026 L\n\n0.4317M = x mols / 0.026 L  \n  x mols = 0.0112242 mols\n\nNow just one last step. Find the molarity of the new solution using same equation. 100 mL = 0.1 L\n\nM = 0.0112242 mols/0.1 L\nYour final M = 0.112242 M
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
895387
NaOH ® Na+ + OH- is sodium hydroxide, a common base.\nWater has equal hydronium and hydroxide ion and is thus neutral as is C2H5OCOOH ® H+ + C2H5OCO O- which is in chemical equilibria and C2H5OH is ethanol that might have a slight alkalinity but is not considered a base.
843345
I will do the Na+ for you. All the others are similarly done.\n\nin 3 M of NaCl you have 3 mol/liter.  In 3 liters you have 9 mol.\nIf you mix both volumes you will have 3 + 2 = 5 liters\n\nso your Na+ concentration will be 9/5 = 1.8M
495941
We have no idea what data you [we] have to work with here.\n\nWhat did you do in your lab?\n\nI assume you had a saturated solution of Ca(OH)2 which you then titrated with a stron acid like HCl.  You then should have recored the volume and concentration of the acid required to neutralize the Calcium Hydroxide.\n\nWrite out the chemical reaction between Ca(OH)2 and your acid, find the mole ratio between the moles of acid used and the moles of OH- in solution, and solve.
588293
When an acid reacts with an alkali (base), neutralisation occurs as the pH of the mixture becomes neutral, pH=7. (Acidic: pH < 7, Basic: pH > 7).\n\nSo an example would be to react sulphuric acid with sodium hydroxide to form sodium sulphate (salt) and water.\n2NaOH + H2SO4 ---> Na2SO4 + 2H20\n\nAn ELEMENT that would neutralise sulphuric acid (i.e. make its pH 7) could be any metal.\n\ne.g. Mg + H2SO4 ---> MgSO4 + H2\n\nBut this is not called a neutralisation reaction.
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
592167
We assume NaOH is a strong base and therefore dissociates completely at 60 degrees C. a concentration of .002M solution of NaOH dissociating completely contributes .002 mols/liter of OH ions to the solution. Since pOH = -log[concentration of OH-], pOH = -log.002 = 2.699 (with correct logarithmic sig figs). pH is 14 - pOH, so the pH of the solution will be 14-2.699 = 11.301.
139469
16.45 mL HNO3 solution * [1 liter/1000 mL] * [0.302 moles HNO3/1 liter HNO3 solution] * [1 mol NaOH/1 mol HNO3] * [1 liter NaOH solution/0.250 mol NaOH] * [1000 mL/1 liter]\n\n=67.0 mL NaOH solution\n\nEDIT: [Sorry, the answer above was wrong; I don't know how I got that. The answer is 19.87 mL]\nEND EDIT\n\nBasically, you just use a whole bunch of ratios relating all the units and measurements together.\nIf you write all that out, all the units should cancel except the "mL NaOH solution"\nMight want to check my answer though...\n\n\nNext question:\n[16.45 ml HNO3 solution] * [1 liter/1000 ml] * [0.302 mol HNO3/1 liter HNO3 solution] * [1000 millimoles solution/1 mol solution]\n\n=4.97 millimoles HNO3 solution
837288
Use the formula M1V1=M2V2\n\nThat is the first Molarity times the first volume are equal to the second Molarity times the second Voume.\n\n.2098M x70mL = .662M x V2\n\nSolve for V2.\n\n(.2098 X 70)/.622 = V2\n\nV2 = 23.61mL NaOH to titrate the solution to neutrality.
117935
I believe that it is 7... If the concentration of the acid and base are equal then they would probably neutralize the solution.
875212
Since HCl and NaOH are both strong solutions (acid and base, respectively), you know the pH is going to be 7.0.
1104578
Use the equation M1V1 = M2V2\n\nM1 is the molarity of the HCL or 9.9 M\n\nV1 is the first volume of 102 ml but it needs to be changed to Liters. So it is 0.102L\n\nV2 is the total volume of the final solution so 525 ml (0.525 L) + V1 or 0.102L = 0.627\n\nM2 is what you are looking for - the final concentration or molarity.\n\nRearrange the equation to ((M1)(V1)) / (V2)  = M2\n\n(0.102L)(9.9M) / (0.627) = 1.61 M\n\nHope this helped!
935131
http://en.wikipedia.org/wiki/Sodium_hydroxide
933864
to what?  Are you looking for the molecular weight of the monoprotic acid?  If so, remember M (molarity) is in moles/lter) and your colume of NaOH is in mL so you'ww want to convert that to liters.\n\n\n.01524 liters x .1578 moles/liter = 0.002405 moles\n\nso therefore you have 3.567  g / 0.002405 moles = 1483 gram/mole\n\nuse dimensional analysis and keep track pf the units to find the answer