Find the amount of organic acid content in a mixture of acid&ester?
610332
carwash's ideas are pretty good, particularly TLC and spectroscopy.\n\nI don't think I'd do a titration against a base, because esters will also react with bases (hydrolysis). For a synthesis, this doesn't matter. But for an analytical method, it invalidates the technique, unless you can account for it with a standard or something. You might end up getting a result higher than what it should be because of it. (It depends on the chemical environment of the ester)
166863
The portion of the vol. of a solution or mixture related to the molar content of one of the components with the solution or mixture.
395302
mixture is combining substances, solution is dissolving a substance in another
1072394
The listed percentage of vinegar is almost always "by weight" and not volume.\n10% vinegar means that 10% of the weight of the vinegar is acetic acid, the other 90% is water.\n\nTo calculate the molality, we need to figure out how many moles of a solute (acetic acid) is dissolved in the solvent (water).\nWe need also calculate the volume of the solution (in liters).\n\nTo solve this, take (for example) a 100 gram sample of 10% acetic acid, this would mean that 10 grams out of 100 are acetic acid.  Knowing the molar mass of acetic acid (60.05 g/mol) we can calculate the number of moles of acetic acid present.\n10 g / 60.05 g/mol = .1665 moles\n\nA 10% acetic acid solution has a density very close to that of water, so it is not unreasonable to just assume a density of 1 g/cm^3 (if you want a more accurate answer, look up the density your self).  If we assume the density is 1 g/cc, then 100 grams of 10% vinegar would have a volume of 100 mL, or .1 Liters.\n\nWe now know the number of moles of solute and number of Liters of solution, we can calculate molarity.\nM = moles / Liter\nM = .1665 moles / .1 Liter\nM = 1.665 Molar
581048
The coefficient before the aluminum is 2 and before hydrochloric acid is 6.  Therefore, there is a 2:6 ratio (or 1:3 if you reduce) between the reactants.  This means for every one mole of Al you will need three times as much HCl.  Thus, if you have 2.4 moles Al, you will need at least 7.2 moles of HCl.
792017
H2C3H2O4 + 2NaOH --> Na2C3H2O4 + 2H2O\n\nmole of NaOH needed = 0.1 * 20.76 = 2.76 mmole\nmole of malonic acid = 1.38 mmole\nmolarity of malonic acid = 1.38 / 13.15 = 0.104 M\n\npeace\nvixklen
624341
A solution that contains an equivalent amount of an acid and its conjugate base.
775430
25% citric acid will have a concentration of 250 g/l.\n\nDividing by 192.1 g/mol we will get:\n\n(250 g/l) / (192.1 g/mol) = 1.3 mol/l = 1.3 M.
946866
11/27:Please post the answer so I can find the error in my work or determine if the answer key is wrong.  Lets see if we can get to the bottom of this one!\n\n\nFirst find the masses of each acid:\n\nHCl:  48.8 ml x 1.19 g/ml x 0.38 = 22.1 g\n\nHNO3:  17.5ml x 1.42 g/ml x 0.70 = 17.4 g\n\nSecond, convert grams to moles:\n\nMass HCl = 35.45 + 1.01 = 36.46 = 36.5 g/mol\nMass HNO3 = 16.0 x3 + 14.0 + 1.0 = 63.0 g/mol\n\nmoles HCl = 22.1 g/36.5 g/mol = 0.605 mol HCl\nmoles HNO3=17.4g/63.0 g/mol = 0.276 mol HNO3\n\nSince you're diluting to one liter, these are also the molarities.\n\n[HCl] = 0.605 mol/L ; [HNO3] = 0.276 mol/L\n\nEach of these acids are strong acids and can be assumed to dissociate completely.  If they were weak acids we would have to account for the acid dissociation equilibria here.  \n\nEach of these acids contribute only 1 equivalent of hydrogen ion.  Thus [H+] is the sum of the two concentrations.\n\n[H+] = [HCl] + [HNO3] = 0.605 + 0.276 = 0.881 mol/L = 0.88 M\n\nReport two sig figs since your mass % vals only have two sig figs.\n\nThe equilibrium equation for water, H+ and OH- is:\n\nH2O <--->  H+  + OH-   K = 1.00 x 10^-14\n\nH2O is a pure liquid and by definition has an activity of unity, so the equilibrium expression is:\n\n[H+][OH-] = 1.00 x 10^-14\n\nNow we can solve for [OH-] :\n\n0.881 X [OH-] = 1.00 x 10^-14\n\n[OH-] = 1.14 x 10^-14 mol/l = 1.1 x 10^-14 M\n\nAgain, two sig figs in final answer.\n\nFinally, pH = - log [H+]\n\nso pH = - log 0.881 = 0.055
1083808
100 mmol/L acetic acid in terms of mass per litre:\n = 100 mmol/L * 60 g/mol = 6000 mg/L or 6g/L which is 0.6% acetic acid.\n\nBut if you start with a 10% solution of acetic acid, it is 10/0.6 or 17 times too concentrated for what you need.  Dilute with 17 volutes of water.\n\nI hope this helps.  There are other ways to show if need be.
1000766
The low technology approach is to use a densitometer.  This can be as simple as a hydrometer as used in the brewing industry to check for specific gravity (density).  From a measured density and a known volume and if you know the densities of both liquid components of a binary mixture you can work out the relative proportions of each using simultaneous equations and hence the concentration of one in the other.
951445
... what's the specific gravity of the mixture?\nSince the total volume is the sum of the two ingredients, there was no volume loss due to mixing. (Sometimes there is; water and ethanol are an example.) Then the total mass is 6*0.8 + 5 * 1 = 9.8 kg and the sp. gr. = 9.8/11 = 0.8909 kg/L.
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
434220
tert-Butyl isobutyrate ~ 143.7g/mol\nethyl isocrotonate = 114g/mole isocroronate acid = 86g/mole, ethyl = 29, -1 for the H on the acid.\n\nPlease check my math
777003
CH3COOH + NaOH ---> NaCH3COO + H2O
384774
CH3CH2OH\n\nC ~ 12\nO ~ 16\nH ~ 1\n\n2*C + 1*O + 6 * H = 2*12 + 1*16 + 6*1 = 46 g/mol
485893
Check on the internet for the ratios used by comercial factories, then prepare a solution from glycol and water in the given proportion. That's all!!!