How many milliliters of a .250 M NAOH solution would it take to react with 16.45 ml of a .302 M HNO3solution?
139469
16.45 mL HNO3 solution * [1 liter/1000 mL] * [0.302 moles HNO3/1 liter HNO3 solution] * [1 mol NaOH/1 mol HNO3] * [1 liter NaOH solution/0.250 mol NaOH] * [1000 mL/1 liter]\n\n=67.0 mL NaOH solution\n\nEDIT: [Sorry, the answer above was wrong; I don't know how I got that. The answer is 19.87 mL]\nEND EDIT\n\nBasically, you just use a whole bunch of ratios relating all the units and measurements together.\nIf you write all that out, all the units should cancel except the "mL NaOH solution"\nMight want to check my answer though...\n\n\nNext question:\n[16.45 ml HNO3 solution] * [1 liter/1000 ml] * [0.302 mol HNO3/1 liter HNO3 solution] * [1000 millimoles solution/1 mol solution]\n\n=4.97 millimoles HNO3 solution
780228
HCL and NaOH are 'strong' acid and base. When they react, the product is neutral. NaOH + HCL --> NaCl + H2O\n\nWhat you need to find is which is the limiting reactant, the other one (what is left over) will effect the pH by its concentration in the otherwise neutral solution.\n\nFor your first problem:\n\n0.0557M HCL = 0.0557mols/liter*0.00542L = 3.02*10^(-4)mol HCL\n\n0.0377M NaOH = 0.0377mols/liter*0.00337L = 1.27*10^(-4)mol NaOH\n\n3.02*10^(-4)mol HCL - 1.27*10^(-4)mol NaOH = 1.75*10^(-4)mols HCL\n\n0.00542L + 0.00337L = .00879L\n\n[(1.75*10^(-4)mols HCL) * (1 L) ] / (.00879L) = .020M HCL\n\npH = -log(.02) = 1.70
901261
I think that the equation is \n3Ba(OH)2 + 2H3PO4 ---> Ba3(PO4)2 + 6H2O\nbut maybe the answer is\nBa(OH)2 + 2H3PO4 ---> Ba(H2PO4)2 + 2H2O\nbut no problem you can use the same  way to calculate.\n\n3Ba(OH)2 + 2H3PO4 ---> Ba3(PO4)2 + 6H2O\nthe moles of H3PO4 0.01015 * 0.141 = 1.43 * 10^-3\nthe moles of Ba(OH)2 needed : 1.43 * 10^-3 *3 / 2 = 2.145 * 10^-3\nthe volume of Ba(OH)2 needed : 2.145 * 10^-3 / 0.0584 = 0..03673 L = 36.73 ml
696206
the moles of NaOH = the moles of OH - = 0.125 * 0.606 = 0.07575\nthe moles of OH - in the new solution = 0.07575/(15 + 0.125 ) = 5*10^(-3) M\npOH = - log ( 5*10^ -3) = 2.3\n=> pH = 14 - pOH = 11.7
829104
Well, it's 0.205 moles/liter and you're making 1/4 liter, so you need 0.205/4 or 0.05125 moles.  The molar mass of NaOH is equal to the sum of the molar masses of Na, O, and H, or 22.990 + 15.999 + 1.007 = 39.996 g/mole.  The mass on NaOH needed is equal to 0.05125mole*39.996 g/mole = 2.050 grams.
646163
Presuming that the base is monobasic.\n\na) with 25 ml added, you are at the equivalence point.  The pH is 7.\nb) after adding 25.01 ml, you will have 0.01 ml of base that did not get reacted.  The molarity of the base in solution will then be 0.1 * (0.01/50.01) = 2.0 x 10^-5.  Your pOH is 4.70, pH would be 9.30\nc) Now, it's 0.1 ml of unreacted base.  Molarity is 0.1 * (0.1/50.1) = 2.0 x 10^-4.  Your pOH is 3.70, pH is 10.30.\nd) Now 1 ml of unreacted base.  Molarity is 0.1 * (1/51) = 2.0 x 10^-3.  pOH is 2.70, pH is 11.30.\ne) Now 25 ml of unreacted base.  Molarity is 0.1 * (25/75) = 3.33 x 10^-2.  pOH is 1.48, pH is 12.52.\n\nIn each case, you have to figure out the amount of unreacted titrant and recalculate the molarity as if you were simply diluting it, using McVc = MdVd, where the volume is now the total volume of the solution (the unreacted amount plus the 25 ml of acid and the 25 ml of base that have neutralized each other).
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)
358982
NaOH is completely dissociated when it is dissolved in water.  A .0025M KOH solution therefore produces a [OH-] equal to .0025M.  The ion product of [H30+] and [OH-] is a constant, 1.0 x 10^-14 M^2.  By substitution, the [H30+] can be determined.  The pH can then be calculated.\n\n[H30+] = 1.0 x 10^-14 M^2/[OH-] = 1.0 x 10^-14 M^2/2.5 x 10^-3 \n\naaaand all that equals .4 x 10^-11... and you put that in correct scientific notation... 4 x 10^-12 M for the concentration of [H30+].\n\nWe then plug that in...   \n\npH= -log[H30+] = -log (4 x 10^-12) = 11.4\n\nThe pH of a .0025M solution of KOH is 11.4, which is greater than 7, making it a basic solution.\n\nStrong base at that.\n\nP.S.- If it is supposed to have the correct number of significant figures/digits, the answer would then be 11.
497859
A)\n\nConvert HCl into millimoles. .2M * 1 mL = .2 mmol. Now, you've diluted it to a volume of 5 L. So you convert back, keeping in mind that you need to divide by mL for the units to cancel:\n\n.2mmol/5000 mL = 4*10^-5 H30+\n\nKnowing that Kw = [H30] * [OH-], and that Kw = 1*10^-14, you can figure out OH- and pOH. (pOH is the -log of OH-)\n\nThe pH of the HCl soln is 4.39, based on -log (4*10^-5)\n\nB)\n\nSodium Oxide, like most alkali metal oxides, reacts with water to create sodium hydroxide an \n\nNa2O + H20 --> 2NaOH \n\nSo it's going to be .2 mol of OH- created, for every .1 mol created. So it's .2 mol NaOH/ 1L = .2M NaOH.\n\nThe OH- is .2M\nThe H+ is figured out using Kw = H+ * OH- (see above)\nThe pOH is 0.70\nThe pH, since pH + pOH = 14 at 25 degrees, is 13.3
820519
First write the balanced chemical equation:\n\nPb(NO3)2 + 2NaBr --->  PbBr2  +  2NaNO3 \n\nThen, find the total volume of the solution: 200mL + 200mL = 400mL \n\nFrom the density, the mass is:   density = mass/volume   mass = density * volume = 1.0 g/mL * 400 mL = 400 g\n\nQ = mC(Tf - Ti) ; where Q = heat released, m = mass of solution,, C = specific heat capacity, and (Tf - Ti) = change in temperature\n\nSo,  Q = 400g * 4.2 J/gK * (2.44 + 273)  (Since C is J/gKelvin, the temperature rise also has to be in Kelvin)\n\nQ = 462.5 KJoules\n\nHope this helps...
808705
Molarity of the solution = Number of moles of citric acid /1000 ml of solution\n\n= (250/192.1) moles of citric acid per 1000 ml of solution\n\n= 1.3014 M
842372
Write the balanced chemical equation:\nNaOH + HNO3 -----> H2O + NaNO3\n\nthis means that 1 mole of NaOH is required to use up 1 mole of HNO3\nthis means that all of your HNO3 will be used up so to determine the excess NaOH all you have to do is subtract the # of moles of HNO3 from the number of moles of NaOH giving 0.00193 moles NaOH left.
318573
The reaction between Hydrochloric acid and Sodium Hydroxide is as follows,\n\nHCl + NaOH --> NaCl + H2O\n\nYou can see that there is a 1:1 ration between the number of moles of HCl needed to completely reacted with NaOH.\n\nSince you are given both a volume and concentration of the NaOH solution, you should be easily able to calculate the number of moles of NaOH availible.\n# mole = volume * concentration\n# moles = .1 Liters * .25 moles/L = .025 moles of NaOH\n\nIf you know you have .025 moles of NaOH, you know you need .025 moles of HCl to react it completely.\n\nIf you know that you have a .15 mole/Liter solution of HCl, you can calculate the volume required to get .025 moles lie this,\nLiters = moles / oncentration\nLiters = .025 moles / .15 moles/Liter = .01667 Liters or about 167 mL of solution.\n\n\nNext time do your own homework.
143234
I think you at least need to know the density of pure acetic acid to solve your problem. We can assume density of water at 1.0 g/mL.\nIf we assume that the density of your 40% w/v solution is 1.0 g/mL then:\n\n    V = volume of 40% solution required would be: \n\n        V x .40  = 0.0225 x 250 \n\n  V = 14.1 mL \n\nYou place the 14.1 mL in a flask and add water to complete 250 mL and voila... You are done. \n\nLet me know if this is wrong.
776525
Since you are pretty, here's how:\n\n1. Compute for the number of moles NaOH needed to prepare the solution.\nmole NaOH: 1 L * 0.5 mol/L = 0.5 mol\n\n2. Get the weight NaOH you need by multiplying the number of moles to the molecular mass of NaOH.\nweight NaOH: 0.5 mol * 40g/mol = 20g\n\n3. To prepare, weigh 20 grams of NaOH pellets. Dilute this with water to 1000 mL in a volumetric flask.
638948
The reaction NaOCl + NH3 = NH2Cl + NaOH is very easy even cold. The reaction NH2Cl + NH3 + NaOH = N2H4 + NaCl + H2O is not so easy as it requires heating to boiling or close to it.\n\nHere's the link: http://www.sciencemadness.org/talk/post.php?action=reply&fid=2&tid=1128&repquote=20454
754271
2H2 + O2 --> 2H2O\nPV = nRT\nP = 2 + 3 = 5\nT = 25 + 273 = 298\n5.00 atm x 55.0 L = n x .08206 x 298\nn = 11.25 mol H2 and O2 mixture\n11.25 mol x (2 mol H2O / 3 mol) = 7.50 mol H2O\n\nP x 55.0 = 7.50 x .08206 x 298\nP = 3.33 atm