A 15.8 g sample of HF is dissolved in H2O to give 200 ml of solution. What is the concentration?
832170
Molarity = moles of solute/liters of solution\n15.8 g of HF = 1.01 + 19 = 20.01 g/mol\n15.8 g HF / 20.01 g/mol HF = 0.790 mols of HF\n0.790 moles of HF / 0.200 L of solution = \n3.95M solution
272078
I take it you mean the concentration of hydronium ions (H3O+) at a pH of 6.2. ok, the equation you need is this:\npH = -log [H+]\n\nTo solve, do this:\npH = -log [H+]\n-pH = log [H+]\n10 ^ (-pH) = [H+]   (if you don't understand this step, look up logarithms. Basically, a= log b means exactly the same thing as 10^(a)=b )\n\n10^(-6.2) = [H+]\nYou have 10^(-6.2) mol/ L of hydronium ions.\n\nAnother useful equation is this:\n[H30+] * [OH-] = 10^-14\nwhich will give you the concentration of OH- ions when you have the concentration of hydronium ions, or vice versa.
224213
well, pH = -log[H+]. so if you just undo the log you will have the concentration of H+ (which is the same as H30+). then .. [H+} + [OH-] = 14 good luck!
575750
This can be done using the formula M1V1=M2V2.  For example, if you have 2M HCl and wanted 2L of .5M HCl, then putting that into this equation as 2(V1)=(.5)(2).  Solve accordingly.  This is just an example.  This formula can be used in a variety of ways that all.  But, to make it to the desired you, a volumetric flask is needed in which water is added until the etched line on the flask to equal the preset volume of the flask.
952215
[H3O+][OH-] = 1.0 x 10^-14 M always at 25C\nso if [H3O+]= 5.0 x 10^-10M\n[OH-]= 2.0x10^-5
586791
MOLARITY=(MOLECULAR WT/GIVEN WT)*(1000/VOLUME OF THE SOLUTION IN MILLIITERS)\n\nBUT WE KNOW THT M1V1=M2V2\nie 35*2.104=M2*825\nM2=0.0896M\nTHE AMOUNT OF WATER ADDED IS 825-35=790.
661226
It doesn't matter how much iron you happen to dissolve as long as you have the exact mass you dissolved for the later calculations. Use just enough sulphuric acid to dissolve all of the iron, then transfer the solution into a measuring bulb and fill up to the mark - the theoretical concentration of iron is the mass you used divided by the volume. \n\nTheoretically you need 1 mol of acid for each mol of iron, but you will want some excess acid to keep the ions in solution. Make sure that oxygen has no chance to react with your solution during dissolving and titration.\n\nTitration with potassium permanganate in acid solution will require 1 mol permanganate for each 5 moles of iron. Usual lab concentrations are 0.02 mol/l (1 ml of that solution detects 0.1 mmol iron). You will want to use roughly 20 ml permanganate solution per titration for a volume error <1%.\n\nNow we start calculating (using SI units). Iron has a molar mass of 55.8... g/mol. If you want to use about 20 ml, you will need about 0.002 mol of iron. \n\nYour iron solution will have a concentration of \n\nC = m / (M * V)\n\nm : mass of (impure) iron you dissolved, in grams\nM : molar mass of iron, in gram/mol\nV : volume of the volumetric bottle you filled your solution up with, in ml (for convenience)\n\nC : concentration of impure iron, in mol/ml\n\nDivide 0.002 by C to see (roughly) how many ml of your iron solution you want for your first titration. Take an appropriate full pipette which fits the order of magniture, or else dilute your original solution in a volumetric bottle so that you can use a full pipette you have at hand.\n\n\nNow to get percentage purity.\n\nI'll assume you can calculate a concentration from titration.\n\nWith your titration you will determine the measured concentration of iron (in small numbers of mol/ml). Divide that result by C (calculated above), and multiply by 100 to get percentiles.\n\nHave fun!
625417
pH= -log(6.0 x10^-3)\npH is about 2.22
168718
The eq constant is 5.286E-16, so this means that 5.287E-16 = [H][OH]. (The [H2O] doesn't matter as it is a pure liquid.) The H and OH concentrations are the same (because of mole ratios), so solve this equation:\n\nx^2 = 5.287E-16\n\nx is the eq. concentration of H, and also the eq. concentration of OH.
946830
Just remember M1*V1=M2*V2 where M is molarity and V is volume. A little Algebraic manipulation gives you:\n\nV1 = (M2*V2)/M1\n     = (0.195*15.5)/18.0\n     = 0.168 L = 168 mL
545577
H2(g)+ O3(g)=> H2O(L) + O2(g)\n\nBecause H2O is the prime factor of 02G if only 2H2 equals 2H20. If you take the "Smithy" equation into consideration and change the X into L factor, you will find that D is the correct answer!
704589
Hydronium ions, in an aqueous solution.  H30+
217873
Measure concentration. Use a dummy (with water only) measuring that with a LabPro probe. Then compare that to a certain volume of a variable soft drink. I did this in high school, hope it works!
838030
If you have 800g of 10% acid water, then you have 720g water and 80g acid.  \nTo have a 16% solution, call the total mass of the solution x.\nThen 16% x is 80g.\n(.16)x=80g\nx=(80/.16)g=500g\nThe total solution is 500g; 80g acid and 420g water,\nso you need to evaporate 300g of water. \n Anmari muzukashikunai deshoo.
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.
904431
Just the take the [H+] ion concentration divided by the initial [HF]:\nfrom your info i gather it should be (10^-1.59)/1.0 \n\nConvert it into a percentage.
472386
(250 X 0.35) + (1350X0.75) + (500X0.15) / (250+1350+500) = \n87.5+1012.5+75/(2100)\n1175 m mol/2100\n1.175 MOLES / 2100ml = 115g of sulfuric acid\n\n1.175moles /2.1 L = 0.58 M SOLUTION