What volume of 3 M NaOH does it take to completely neutralize 700 mL of 6 M HCl?
354896
1400 mL of NaOH needed...\n\nneed to make sure the concentrations 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
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.
318519
One mole of NaOH weighs 40gms. 200mmol is 0.2 of that and 100ml is 0.1 of a litre. \nSo you would need 40 X 0.2 X 0.1gms of NaOH.=0.8gms\n\nIn reality you would likely dissolve 8gms of NaOH into a 1L volumetric flask and fill it to the mark. You would then measure out 100 mls of this solution. Weighing out 0.8 gms of NaOH is inaccurate because it reacts with CO2 in the air and absorbs water vapour.
474679
0.035=7/200 therefore for every 7 parts Br2 you must have 200 CCl4. (200/7)*50 = 1428.6mL
9807
Well lots of chemistry questions lately....finals coming up? :)  Good Luck!!!  \n\nTo calculate this type of question first of all calculate the #of moles of HCl used.  So \nyou used 2100 mL = 2.1 L HCl \nThis HCl is 0.3 mol/L so you used:\n\n2.1 L HCl* (0.3 mol HCl/L) = 0.63 mol HCl\n\nnow use the ratio of Cl/HCl  (which is 1 to 1)\n\n0.63 mol HCL *(1mol Cl/1mol HCl)= 0.63 mol Cl ion\n\nSo far so good I hope.  \n\nNow do your solution diltion to calculate volume,\n2.1L + .2385L = 2.3385 L of mixed solution\n\nNow simply divide \n0.63 mol Cl/2.3385 L solution = 0.269 mol Cl ion in final solution.\n\nAgain these numbers are quick and dirty but the method of solving this type of problem is the same.
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.
424592
H2SO4 + 2NaOH ----> 2H2O + Na2SO4\n\nYOU MUST SEPARATE AND CANCEL THE COMMON IONS (AQUEOUS ONES)
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
1048754
it depends on NaOH consentration\nit must be more than 7\nthis formula may help you\npH = 14 + log [NaOH]\n14 figure come from negative logatherm of Kw, 10^-14\n[NaOH]=NaOH concentration in mole/L
910262
There is no sodium hydroxide. One mole of H2SO4 will react with 2 moles of NaOH in a neutralization reaction.
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
527028
This time i ans ur question but next time plz do ur homewrk ur self.\n\n1.\nWeight of NaOH = (4*.05) = .2 grams\nMoles of HCl =  Moles of NaOH = (.2/40) = .005\nVol. of HCl = .005*22.4 = .112 l = 112 ml\n\n2.\nWell let the grams be x\n1000x = 74.5*2.25*(75+x)\n(1000-167.25)x = 167.25*75\nx = 14.98 grams
706503
find the heat evolved, assuming heat cap of water is 4.2J/g/K\n\n(31.43-25.32) x 4.2 x 100g = 2566.2J\n\nnow according to definition, molar heat of neutralisation is calculated upon the mols of water formed. this number of mols is equivalent to the no. of mols of the limiting reagent (in this case, the HCl).\n\nmol of HCl= 0.050 x 1 = 0.05mol\n\nenthalpy of neutralisation= 2566.2/0.05= 51324J/mol\n\nthis answer is pretty close to real-life experimental values for strong acid-strong base.
573222
Once a solution of HCl hits pH 7, it is considered neutralized (or in other words it'll basically be water.) After a pH of 7, a solution has to be alkaline (or basic,) which an acid cannot be.
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
432315
You'll need to give the degree of dissociation of the acid CH3COOH. If it was a strong acid you have to used the formula \nMV/mv = 1   where M is molarity of NaOH,V is its volume\n            and m is molarity of CH3COOH, v is its volume\n\nActually the problem here is CH3COOH is a weak acid and it would not have dissociated fully, hence the above formula cannot be used.
600936
Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O\n\nno. of mol of Na2CO3 reacted\n= m/M\n= 1.5/106.2\n= 0.01412 mol\n\nno. of mol of HCl reacted\n= 95/1000 x 6\n= 0.57 mol\n\nsince Na2CO3 is the limiting reagent, the no. of mol of CO2 liberated = 0.01412 mol\n\nMass of CO2 liberated\n= 0.01412 x 44.01\n= 0.6214g\n\nVol. of CO2 gas \n= 0.6214 x 1.96\n= 1.217\n= 1.22g (3s.f.)