Determine how many moles of H30+ ions would be present in 1 litre of 0.1M sulfuric acid solution.?
611049
pervious answer is incorrent.\n\nThere are 0.1 moles of H30+ ions in the solution as once sulphuric acid has dissociated once it creates the rather stable conjugate base HSO4+ which is what makes sulphuric acid "strong".  This ion has a very small pK value, and is to all purposes not dissociated.  Reaction of the solution with a base will encourage it to dissociate tho.
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!
854886
1. 1,000  (each full number in the pOH scale is a factor of ten in the concentration).\n\n2.  C, because you have about 98 grams of H2SO4\n\n3.  b is an acid and its salt.\n\n4.  d because you have about five moles of water.
770958
1/1000 x molar mass (grams/litre) and solute in 1 litre
1108880
HCl   +  NaOH   ===   NaCl   +   H2O\n0.0015 ,  0.0016 ,             0        ,       0        at start of reaction (moles)\n0,           0.0001 ,            0.0015 ,    0.0015  at end of reaction(moles)\n\nevery mole of water contributes about 10^-7(ten to the power -7)\nmoles of OH-  and H+ . thus their contribution to the pH is neglected when there is a considerable amount of acid or base present  i.e more than 10^-5 moles/litre (not moles but CONCENTRATION)\nThus concentration of NaOH = 0.0001/90*1000 moles/litre{total volume = 50+40 ml} = 0.001 .\nas NaOH is a strong electrolyte like any other sodium compound  we can assume complete ionisation . thus[OH-]=[NaOH] = 0.001. now pH = -log[H+]=14-(-log[OH-])= 14+log[OH-]  =14+log[0.001]=11\nthus the pH=11
487714
assuming the equation\n\n2HCl + Mg --> Mg2+ + H2 +2Cl-\n\ntake the moles of Mg you have... multiply by 2 and that is how many Moles of HCl you need. multiply by 2 because one Mg reacts with 2 HCl's so you need twice as much HCl. now that you have moles of HCl convert to volume of HCl, i.e. divide moles / concentration (in this case 2.O M) and that will give you liters of HCl needed.\n\n(.000382 moles *2)/2 = .000382 L = .382mL
939677
Long answer:  Use the vol. and concentration of NaOH to figure out how many moles of NaOH you have which will = moles of OH (since NaOH has only one OH).  Formula is:\n\nMolarity = moles/liters (make sure the volume is in liters)\n\nIn a neutralization reaction the moles of OH = moles of H+\n\nOnce you have moles of H+, use that answer and the volume of HCl to calculate concentration of H+ which will be = to concentration of HCl since HCl has only one H+\n\nUse the same Molarity formula\n\nShort answer:  As long as the base (NaOH) and the acid (HCl) have only one OH and one H, the short form is:\n\nMb x Vb = Ma x Va\n\nwhere Mb = molarity of the base\nwhere Vb = volume of the base\nwhere Ma = molarity of the acid\nwhere Va = volume of the acid
968497
The initial concentration of the acid is C=mole/V=1/2=0.5 mole/L\nLet's assume that the acid is HA\n\n.. .. .. .. .. .. .. HA <=> H(+) + A(-)\nInitial .. .. .. .. 0.5\nDissociate .. .x\nProduce .. .. .. .. .. .. .. x .. .. .. x\nAt Equil. .. . 0.5-x .. .. . x .. .. .. x\n\nKa=[H+][A-]/[HA] =x^2/(0.5-x)\n\nbut x=[H+]=3.4*10^-2 so\n\nKa= ((3.4*10^-2)^2) / (0.5-3.4*10^-2) =2.48*10^-3
844407
Easy, the molecular weight of HCl is 36.461 gr per mole. \n\nIf your trying to reach 4 molar solution then you need 4 x 36.461g HCL diluted to 1 liter. you need 145.844g HCl in the liter\n\nBut since your probably using 32-38 % HCl (actual amount might be on the bottle) you need to multiply you'll need to adjust for inactives by assuming your active molar mass is ~35% of 36.461 or 12.761g \n\nin order to make 1 liter of 4 molar solution you'll need 416.71g \n\n36.461/12.761 * 145.844 = 416.71\n\nassuming it is 35 % active. \n\nYou should always standardize your solution after mixing to determine final activity!!!!!
379843
Using the formula for pH: pH = -log(H+)\nwe can find the molarity\n\nFirst, we plug in what we know\nThe pH is 2.65\nand the H+ is the concentration of hydrogen ions measured in moles/ liter. Hey, that's what we're looking for!\n\n2.65 = -log(moles/liter)\n\nusing algebra we can solve for moles/liter\n\n-2.65 = log(moles/liter)\n\nLet's raise 10 to the power of each side\n\n10^(-2.65) = 10 raised to the negative 2.65 power\n10^(log[moles/liter]) = 10 raised to the power of log(...)\n\nOk, so..\n\n10^(-2.65) =10^(log[moles/liter])\n\nNow 10 with log as a power cancel each other out so...\n\n10^(-2.65) = moles/liter\n\nmoles/liter is the molarity, let's call it "M"\n\nM = 10^(-2.65) \n\nM = approx. 3.16 x 10^(-3)
1006885
I think you are leaving out the first step.\n\nRemember that 12.0 M means 12 moles of HCL per liter of aqueous solution.\n\nTo understand the equations so that you can use them correctly, first think that you will have to add some amount of water to some amount of the concentrated solution to make it the dilute solution.\n\nso, 12 mol/liter , assuming you had a liter of the solution, you keep the same number of moles and add water\n\nhow much water do you add to get to the weak, .6 mol/liter concentration\n\n12 mol/ (1+x) liters = .6 mol/liter\n\n12/(1+x)=.6\n12=.6(1+x)=.6+.6x\n11.4=.6x\n19 = x\n\nso if you had a liter of the solution at 12 M, you would have to add 19 liters of water and then you would have 20 liters of solution at the diluted .6 M\n\nbut you don't need that much of the final solution, so you use the same ratio of water addition, but only start with enough concentrated to end with 300 ml\n\nthe ratio is 19 to 1\n\nso 1/20 * 300 = 15   you need 15 ml of the conc. solution\n   19/20 * 300 = 285  you add 285 ml of water \n\nthe total is 300 ml of the .6 M solution\n\nnow look back at the equations and you will see where they came from and how to apply them properly
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.
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
853370
The concept of 1.5M hydrogen gas is rather odd, but, yes, you use the Nernst Equation.  Don't forget to convert the pH to a molarity of H+ ions.
813197
0.0821 moleP1V1/T1=P2V2/T2\n375*4/293=760*V2/273.5\nV2=375*4*273.5/(293*760)\n22.4 litres contains 1mole\n375*4*273/293*760\n=375*4*273/293*760*22.4\n=0.0821 mole
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! ) :)
502184
5H2O2(aq)+2KMnO4(aq)+\n3H2SO4(aq)-->\n2MnSO4(aq)+K2SO4(aq)+\n8H2O(l)+5O2(g)