How can I find the milliliters of a base used to titrate an acid if it is not a neutralization reaction?
817465
You know three of the four things needed. Try C1V1=C2V2, or an equality.\n\n                          m/ml=m/x
350659
Why do people answer complete and total bullcrap answers?\n\n\nThe only thing they were right about was it is a weak acid, and even that bears no relavance to its P.H only you use a diferent formula in calculating it\n\n\nthe question is incomplete, you need to know the molarity of the acid and then you get the square root of it disociation constant (Ka) multiplied by its molarity\n\nand then you -log(base10)the answer you got from the first part and that is the P.H\n\nin short  Sq root(Ka*molarity acid)= H+ concentration\n\nP.H= -logH+
575881
you can't do this question without knowing the mole ratio of the base and acid. \n\nassuming the number of moles of the acid is the same and the number of moles of the base is the same: \nn(acid) = n(NaOH)\nm/M = CV\n0.55/M = 0.240 x 0.04011\nM = 57.13454...\nM = 57.13 (to d.p.)
979086
Your question is incomplete. With these amount of info you can't find either.\n\nThe equivalent weight is MW/a where a is the valency (for acid/ bases number of H+/OH- ,for redox reactions number of electrons gained/lost). If you don't know the value of a you cannot find the MW from the EW.\n\nFrom the info you have you can find the EW (if you know the reaction for the titration).\nJust as mole=mass/MW, greq= mass/EW.\nIn the titration you have greq Ca3(PO4)2= greq unknown\nbut greq Ca3(PO4)2= a1*mole1= a1*mass1/MW1\nso\n\na1*mass1/MW1=mass2/EW2 =>\n\nEW2=(mass2/mass1)*(MW1/a1)\n\nMW1=310 (for Ca3(PO4)2)\nmass1= 50.5 g of Ca3(PO4)2\nmass2= 34.37 g unknown\n\na1=? You don't specify the reaction so we don't know how much is a1. If you just reach the first equivalence point (phosphate acting as a base) a1=1, if you reach the second a1=2 and if you reach the third a1=3. It could also be a precipitation titration...
808549
M=molesof solute/Liter of solution\n\nM= molarity
518068
Make a solution saturated with calcium hydroxide.  If there is precipitate calcium hydroxide then that means there is as much dissolved calcium hydroxide as possible.  Filter the solution to get rid of the solid.  Now you have a saturated solution, and your objective is to determine the molarity of this solution.  Measure the volume and keep track of this measurement.  To determine the number of moles of Ca(OH)2, titrate it with strong acid like HCl, and see when the pH reaches 7.  When you titrate, keep track of the amound of acid you are adding, and the current pH of the solution.  At pH=7, the number of moles of HCl you added is equal to the 1/2 the number of moles of Ca(OH)2 (because the molarity of H+ and OH- are equal).  Take this number of moles and divide it by the original volume.  This value is the concentration of hydroxide [OH-].  Half of this value is the concentration of calcium [Ca++].  Ksp = [Ca++]·[OH-]²
839352
If tha atomic mass of M is A, then the molecular weight is MW=2*A+60\n\nThe concentration of the M2CO3 solution is C=mole/V and since mole=mass/MW, C=mass/(MW*V)\n\nThen you do a titration for which we assume complete reaction and not reaching only the first end-point, so the reaction is\n\nM2CO3 + 2HCl -> 2MCl + CO2 + H2O\n\nTherefore 2*mole M2CO3=mole HCl =>\n2*C1*V1= C2*V2=>\n2*V1*mass/(MW*V) = C2*V2 =>\nMW= 2*V1*mass/(C2*V2*V)=\n=2*0.025*7.46/(0.1*0.027*1)\n=138\n\nand MW=2*A+60=138=> 2A=78 => A=39\n\nSo M, is potassium (K)
977807
M = moles/L\nso, 0.220M = 0.220 moles per liter and there are 2 liters so there are...\n0.440 moles of Mg(NO3)2\n\nBUT... notice that there are 2 ions of NO3 for each molecule, so double it again and get 0.880 moles of NO3
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
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
743363
1. Get the number of moles of carminic acid.\nsince 1 mole of NaOH neutralizes 1 mole carminic acid, number of moles NaOH equals number of moles carminic acid:  \n0.01802L * 0.0406mol/L = 7.3 x 10^-4 mol\n\n2. Determine the molar mass of carminic acid.\nSince you are given the weight of carminic acid, you can get the molar mass by simply dividing it with is number of moles from #1.\n0.3602g / 7.3 x 10^-4 mol = 492.34 g/mol\n\n3. Calculate for the molecular formula.\na) Assume 100g carminic acid such that weights of C, H, and O are as follows: C = 53.66g; H = 4.09g; and the remaining \nO = 42.25g\n\nb) Get the number of moles of each element by dividing the weights by atomic mass. \nC => 53.66g/ 12g/mol = 4.47mol\nH => 4.09g/ 1g/mol = 4.09mol\nO => 42.25g/ 16g/mol = 2.64mol\n\nThis is your empirical formula: C4.47H4.09O2.64\n\nc) To get the molecular formula, the molecular mass must be equal to 492.34 g/mol. In this case, you have to multiply the empirical formula by ~5 to get C22H20O13. To check,\nC = 22* 12g/mol = 264g/mol\nH = 20* 1g/mol = 20g/mol\nO = 13* 16g/mol = 208g/mol\nwhen you add all up, you get 492g/mol.\n\nSo, molecular formula for carminic acid is C22H20O13.
574036
The first step, find the moles of both.\nYou should get 25 mmol benzoic acid, and 15.6 mmol NaOH.\n\nThen you must find after the reactions, how many moles remain.  \nYou should get 9.4 mmol benzoic acid, 0 moles NaOH, and 15.4 mmol benzoate.\n\nYou have some benzoic acid and some benzoate ion, a conjugate acid/base pair.  This is a buffer solution, so use the Henderson-Hesselbach equation.  You should get pH = 4.4.
729647
Technically with molar amounts being equal, an acid and a base react in what is called a "neutralization reaction"...\n\nie:\n\nHCl + NaOH ---> NaCl and water\n\nBasically you get a salt and a water.
602416
http://www.scienceaid.co.uk/titration.html
266266
Sure.\n\nConvert into moles.\n\n0.10mol/L HCl * 25mL = 2.5mmol = 2.5e-3 mol HCl\n\nSo, you need 2.5e-3 mols OH- to neutralize the H.\n\n2.5e-3 mols OH- * (1 mol Mg(OH)2/2 mol OH-) = 1.25e-3 mols Mg(OH)2.\n\nConvert into grams using the molar mass.\n\n1.25e-3 mols Mg(OH)2 * 58.33 g/mol Mg(OH)2 = 0.0729g Mg(OH)2
940878
A titration is a method of analysis that will allow you to determine the precise endpoint of a reaction and therefore the precise quantity of reactant in the titration flask. A buret is used to deliver the second reactant to the flask and an indicator or pH Meter is used to detect the endpoint of the reaction.\n\nDoing a Titration \n   \nBegin by preparing your buret, as described on the buret page. Your buret should be conditioned and filled with titrant solution. You should check for air bubbles and leaks, before proceding with the titration.\n  \nTake an initial volume reading and record it in your notebook. Before beginning a titration, you should always calculate the expected endpoint volume.\n\nPrepare the solution to be analyzed by placing it in a clean Erlenmeyer flask or beaker. If your sample is a solid, make sure it is completely dissoloved. Put a magnetic stirrer in the flask and add indicator.\n\nUse the buret to deliver a stream of titrant to within a couple of mL of your expected endpoint. You will see the indicator change color when the titrant hits the solution in the flask, but the color change disappears upon stirring.\n\nApproach the endpoint more slowly and watch the color of your flask carefully. Use a wash bottle to rinse the sides of the flask and the tip of the buret, to be sure all titrant is mixed in the flask.\n  \nAs you approach the endpoint, you may need to add a partial drop of titrant. You can do this with a rapid spin of a teflon stopcock or by partially opening the stopcock and rinsing the partial drop into the flask with a wash bottle. Ask your TA to demonstrate these techniques for you, in the lab.\n  \nMake sure you know what the endpoint should look like. For phenolphthalein, the endpoint is the first permanent pale pink. The pale pink fades in 10 to 20 minutes.\n\nIf you think you might have reached the endpoint, you can record the volume reading and add another partial drop. Sometimes it is easier to tell when you have gone past the endpoint.   \n \nWhen you have reached the endpoint, read the final volume in the buret and record it in your notebook.\n  \nSubtract the initial volume to determine the amount of titrant delivered. Use this, the concentration of the titrant, and the stoichiometry of the titration reaction to calculate the number of moles of reactant in your analyte solution. \n  \nTitrating with a pH meter \n   \nTitration with a pH meter follows the same procedure as a titration with an indicator, except that the endpoint is detected by a rapid change in pH, rather than the color change of an indicator.\n  \n   \nArrange the sample, stirrer, buret, and pH meter electrode so that you can read the pH and operate the buret with ease.\n   \nTo detect the endpoint accurately, record pH vs. volume of titrant added and plot the titration curve as you titrate.
861217
molarity = moles/L\nso just plug in what you have!  6.3M= mol/0.216 L   just multiple 6.3 by 0.216 and you will get the moles of the acid (i dnt have a calculator on me right now, but just do the calc out and theres ur answer! ) :)