What vol of 0.200 M HCl sol is needed to neutralize 25.0 mL of 0.150 M NaOH sol?
371296
well seeing that the molarity of the two solution is a direct ratio of how many molecules there are in the solution, i think that the ration would be 0.15/0.20\n\nso assuming that the reation is 1 to 1 (i cant remember so if it isnt 1 to 1, then i'm wrong) then you would need 18.75 mL of the 0.200 M HCl
341279
125mL of 0.251 M HCl gives 0.125 * 0.251 mols of HCl\n\nThen 250 + 125 mL is 375 mL = 0.375 L\n\nSo 0.125 * 0.251 mols in 0.375 L gives 0.125 * 0.251/0.375 M\n\nThat is 0.084 M
745017
Your question isn't really complete - the NaOH needs someting to react with.\n\nFor example:\n\nNaOH is used with lipids to make soap.\n\nNaOH + HCl -> H2O + NaCl
302581
add 2 parts HCl 1 part H20 to dilute properly
833268
I'll tell you how to figure this one out...\n\nStart by always converting to moles...\nConvert 25g of LiOH to moles using its molecular weight.\n\nOnce you have that, you set up dimensional analysis to make your units cancel...\n\n0.139M is the same as 0.139 mol/L \n\nso... your answer above in mol * (L/0.139 mol) * (1000mL/L) = your final answer... the units all cancel and you are left with mL
952447
I think you probably mean that the half liter is mixed with 250 ml. Otherwise the volume afterwards couldn't be 750 ml. Okay, the amount of HCl in the half liter is\nn=V*c=0.5l*2.5 mole/l=1.25 mole (remember that M simply means mole per liter).\nThe amount of HCl in the 250 ml is\nn=V*c=0.25l*3.75 mole/l=0.9375 mole\nThe total amount of HCl is then 1.25 mole+0.9375 mole=2.1875 mole meaning that the concentration in the final solution is\nc=n/V=2.1875 mole/0.750 l=2.92 mole/L=2.92 M.\n\nSince HCl is a strong acid it is completely separated in H+ and Cl- in the solution meaning that\npH=-log(concentration of H+)=-log(concentration of HCl)=-log(2.92)=-0.5.\n(Strictly speaking this formula for pH is only valid in dillute solutions like 0.1 M and below, but it is the best estimate we can give without making this problem VERY compliated).
356281
MgCO3 + 2HCl ---> MgCl2 + CO2 + H2O\n\n2HCl + MgO ---> H2O + MgCl2\n\nBarium hydroxide is a base so can be neutralised with any acid.  An example being HCl.\n\nBa(OH)2 + 2HCl ---> BaCl2 + 2H2O\n\nor H2SO4\n\nBa(OH)2 + H2SO4 ---> BaSO4 + 2H2O
961959
0,220M/L*2,20L= 0,484 moles of HCl\n12,00M = 0,484 moles/?\n? = 0,0403 L or 40,3 mL of 12,00 M HCl solution + 2,16 L of pure water, so you can get you 2,20 L of 0,220 M HCl solution
658363
This is a limiting reagent problem. The first job is to figure out how many moles of each reagent you have and then figure out which one will run out first because that will determine how much product you can make. It is easiest to write out the balanced equation first to see what you are dealing with:\n\n2Na + Cl2 --> 2NaCl\n\nYou have 23g x 1mol/23g the molar mass of sodium = 1 mol of sodium\n\nYou plug the standard pressure, 1 atm, the volume, 0.56 L, the R constant, 0.082, and the standard temperature, 273 K, into the ideal gas law formula PV = nRT and solve for n.\n\nYou get 0.0250 moles chlorine gas.\n\nThen figure out how many moles of chlorine are needed for the amount of sodium you have.\n\nfor 1 mol of Na you would need 1 x 1/2 = 0.5 moles of chlorine. But you only have 0.025 moles so this must be the limited reagent.\n\nNow you must figure out how many moles of NaCl form from 0.025 moles of Cl2.\n\n0.025 x 2/1 = 0.05 moles of NaCl.\n\nNow multiply this by the molecular weight for NaCl, 23 g/mol for Na and 35.5 g/mol for Cl = 58.5 g/mol total\n\n0.05 mol x 58.8 g/mol = 2.93 or about 3 grams of NaCl formed
777607
Since they are both strong acids and bases you can assume: \n\nMaVa = MbVb\n\n(25/1000) * (1.2) = Mols NaOH\n\nconvert Mols to grams and that's your answer
239849
.7M x 20mL = .25M x (Volume of new IA solution)\nVol = (.7 x 20) / .25\nVol= 56mL
685497
Molarity = moles/volume, so volume = moles/molarity\n\n18.5 grams NaOH /40.0 g/mol = 0.4625 moles\n\nVolume = 0.4625/0.750 = 0.6167 L = 616.7 ml
939677
Long answer:  Use the vol. and concentration of NaOH to figure out how many moles of NaOH you have which will = moles of OH (since NaOH has only one OH).  Formula is:\n\nMolarity = moles/liters (make sure the volume is in liters)\n\nIn a neutralization reaction the moles of OH = moles of H+\n\nOnce you have moles of H+, use that answer and the volume of HCl to calculate concentration of H+ which will be = to concentration of HCl since HCl has only one H+\n\nUse the same Molarity formula\n\nShort answer:  As long as the base (NaOH) and the acid (HCl) have only one OH and one H, the short form is:\n\nMb x Vb = Ma x Va\n\nwhere Mb = molarity of the base\nwhere Vb = volume of the base\nwhere Ma = molarity of the acid\nwhere Va = volume of the acid
737901
The question is not worded correctly.  Rewrite the question and I'll be glad to help you.  \n\nYou are asking for the Concentration of NaOH, but it is already given as .10 M as written in your question above.\n\nTo solve a normal titration problem you need 3 of the 4 variables below:\n\nconcentration of base\nvolume of base\nconcentration of acid\nvolume of acid\n\nLet me know.
779566
17.13 mL = 0.01713 L\n\n0.01713 L x 0.3150 M = 0.005396 mol\n\nNaOH is a strong base, HCl is a strong acid.\nEach NaOH releases one hydroxide anion.\nEach HCl releases one proton.\n\nTherefore it takes 0.005396 mol of H+ to neutralize 0.005396 mol of -OH.\n\n0.005396 mol / 0.025 L = 0.2158 M\n\nThe concentration of the NaOH solution is 0.2158 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
1145574
0.150 M NaOH\n\n0.150mol / L  X  40 g / mol  X  1 L / 1000 mL  X  200 mL  =  1.2 grams of solute (NaOH)