A 51.24g sample of Ba(OH)_2 is dissolved in enough water to make 1.20 liters of solution. How many mL of this solution must be diluted with water in order to make 1.00L of 0.100 moles Ba(OH)_2?
207379
start with computation of molarity of solution:\nM_BaOH = (51.24g/ MW BAOH)/ 1.2 L\n\nthen \nusing M1V1 = M2V2\nM_BaOH (x) = (0.1) (1L)\nsolve for x
525814
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
397107
2NaOH + H2SO4  = Na2SO4 +2H2O                                      Sodium Hydroxide is 0.5moles.Two Moles neutralise One mole of Acid.   0.5Moles neutralise of Alkali neutralises 0.25Moles of Acid                                                                                           50mls of 1M SULPHURIC ACID = .05M THEREFORE MOLAR CONCENTRATION IS 5
406652
Molarity (M) = no. of moles(n)/ Volume of the solution(V) in litres\ni.e., M=n / V\n\nnow u need 3 moles\nHence n=3\nfrom 2 M solution\nTherefore M =2\n\nNow\nV = n / M = 3 / 2 = 1.5 L
457927
Molarity is moles per liters:  1.00 mol/1.00 L.  \n\nSo, you will put the number of moles over /0.25L and solve.\n\nThe key here is to know how many grams of NaOH are in a mole.\nYou should have a chemical index of some kind to tell you....\n\n(It's 40.0.)\n\nSo, you have 12.0/40.0 moles of NaOH, or 0.3 moles.\n\nPlugging that back into your formula, 0.3/0.250 = a molarity of 1.20.
415680
1) Calculate the mass of acid in 1 liter:\nIn one liter we have 0.01 mol = 1.321 gram\n\n2) Calculate the dilution factor:\nWe have to dilute by a factor 5/1.321 = 3.785..  \n\n3) Calculate the volume of the 5g/litre solution we have to dilute to obtain 100 ml:\nAs the dilution factor was 3.785, we need 100/3.785 = 26.42 ml of the original solution.
350524
ANSWER: 0.482 M\n\n\nEXPLANATION:\n\nM = mol solute / L soln\n\nTherefore, you need to find the mol of solute and L of soln to find M.  \n\nTo find mol solute:\n10800mg Na(1g Na/1000mg Na)(1mol Na/22.99g Na) = 0.4698mol Na\n\nTo find L of soln:\n1kg soln(1000g soln/1kg soln)(1ml soln/1.025g soln)(1L soln/1000ml soln) = 0.9756L\n\nUsing the two values found plug in to M = mol solute / L soln\n\n0.4698mol Na / 0.9756L soln = 0.482 M
422184
Yes, you are correct.\nEveryone else who has posted so far is mistaken.\n\nGiven:\nM = mol of solute / liters of solution.\n\nFirst Step:\n\nSolving for x (liters of solution), you get:\n.40 / .096 = x\nx = 4.167 L\n\nSecond step.\nL => mL\n4.167 / 1000 = 4,167 ml or app. 4000 mL\n\nYour answer is approx. correct because you were asked for mL of water, which you found, not L of solution, which everyone else seemed to think.  You may lose points for not solving to the correct amount of significant figures though.
478861
Well, in pure water, if you let x = [Mg2+] the [OH-] = 2x.  Per the equilibrium expression 5.61 x 10E-11 = 4xE3 so x=2.4 X 10E-4 M\n\nIn the NaOH solution, the [OH-] is 0.180 M (the contribution from the Mg(OH)2 is negligible).  So in this case\n                5.61 x 10E-11 = [Mg](.180)E2  \nSolving for [Mg], I get 1.7 X 10E-9 M\n\nCheck my arithmetic, I often get careless punching the calculator.
525780
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
7139
We will assume that 2.0 M NaCL(aq) is a dilute solution.  Therefore, 1 L of water weighs 1 kg.  For concentrated solutions, such as acids, you need to know the density of the solution.\n\nTo know the boiling point, use the formula\nDT = Kb * m\n\nWhere DT = change in temperature\nKb = boiling point constant (0.52 for water) \nm = molality.\n\nFirst, calculate the molality of the the solution\n\nMolality = moles of solute per kiograms of solven\n\n\n2.0 moles of NaCl / 1 kg solvent = 2.0\n\nDT=Kb * m\n= .52 * 2.0\n= 1.04 degrees\n= 1.0 degrees (using 2 significant digits)\n\nSo, water boils at 100 degrees C, so the boiling point of 2.0M NaCl is 101 degrees C.\n\nHope it helps.\n\n-Mark
403285
Here's what you do:\n\n1) Write the balanced equation for the reaction.  How many moles of BaSO4 does each mole of BaX2 create? \n2) Determine the F.W. for each BaX2, and for BaSO4 (by the way, the unknown won't be BaF2. It's insoluble: http://www.ktf-split.hr/periodni/en/abc/kpt.html ) Determine the number of moles there are in 0.0944 g of BaSO4, and in 0.158 g of each BaX2.\n3) For which BaX2 does 0.158 g represent the the correct number of moles?  Note: I worked this out, and got a number that was very close, but not an exact match.  That's realistic, since the BaX2 would not be absolutely pure, SOME minute amount of BaSO4 will dissolve, etc.
383120
1. Make sure that HCl is standardized.\n2. Write the balance equation:\n\nCa(OH)2 + 2HCl ---> CaCl2 + 2H2O\n\n- this means that 1 mole of Ca(OH)2 reacts for every 2 moles of HCl\n\n3. Get the number of moles of HCl reacted by multiplying the volume used to react Ca(OH)2 by the actual concentration of HCl (value from standardization). Divide the result by 2 - and this is now your moles Ca(OH)2.\n\n4. Get the molar solubility of Ca(OH)2 by dividing its number of moles by its volume. Unit is in gram per liter.
139178
The amount is diluted by a factor of 25.  So the resulting solution would have a concentration of 11.6M divided by 25 or \nor 0.464M.
481031
Okay first, you have to deduce how many moles of acid react with the base.\n\n1 mol HCl ----------> 1 Mol NaOH\n\n1 H+ --------> 1 OH-\n\nTherefore:\n\nMolarity Acid X Volume Acid = Molarity Base X Volume Base\n\nMaVa = MbVb\nMa(20cm3) = 1.2 X 25 cm3\nMa =  30/20\nMa = 1.5 moles per dm-3\n\nThat gives the MOLARITY of the acid. The concentration is in grams per dm-3. \n\n1.5 moles of HCl = (Molar Mass) X 1.5\n                            = 36.46 X 1.5\n                            = 54.69g\n\nSo effectively;\n\nThe Molarity = 1.5 moles dm-3\nThe Concentration = 54.69 grams per dm-3\n\nWhich ever you need!
74400
A quantity of 22% acid solution is being diluted with some 8% acid solution. The final solution must be 12% acid & contain 101 mL. How much of each solution must be added to accomplish this?- write and solve a system of equations that describes this.\n\nx = amount of the 22% soln\ny = amount of the 8% soln\nx + y = amount of the 12% soln = 101 ml\n\n.22x + .08y = (x + y)*(.12)\n.22x + .08y = .12x + .12y\n.10x - .04y = 0\n\nx + y = 101\nx = 101 - y\n\n.10(101 - y) - .04y = 0\n10.1 - .10y - .04y = 0\n10.1 - .14y = 0\n10.1 = .14y\n(10.1)/(.14) = y\n72.143 ml = y\n\nx + y = 101 ml\nx = 101 - y\nx = 101 - 72.143\nx = 28.857 ml
529919
I think you probably mean that the half liter is mixed with 250 ml. Otherwise the volume afterwards couldn't be 750 ml. Okay, the amount of HCl in the half liter is\nn=V*c=0.5l*2.5 mole/l=1.25 mole (remember that M simply means mole per liter).\nThe amount of HCl in the 250 ml is\nn=V*c=0.25l*3.75 mole/l=0.9375 mole\nThe total amount of HCl is then 1.25 mole+0.9375 mole=2.1875 mole meaning that the concentration in the final solution is\nc=n/V=2.1875 mole/0.750 l=2.92 mole/L=2.92 M.\n\nSince HCl is a strong acid it is completely separated in H+ and Cl- in the solution meaning that\npH=-log(concentration of H+)=-log(concentration of HCl)=-log(2.92)=-0.5.\n(Strictly speaking this formula for pH is only valid in dillute solutions like 0.1 M and below, but it is the best estimate we can give without making this problem VERY compliated).