After 10.0, 20.0 and 30.0 mL of base has been added.\nThe answers are 4.74, 8.75, and 12.36 respectively but I do not know how they reached this answer. Thanks!
136809
You sure did pick a really long question for explaining.  And I'll have to argue that all of the pH's do not have correct sig figs but the numbers are right.\n\n(Sorry about the numbers in the tables but this place doesn't seem to allow better tables)\n\nA) 25.0mL of HAc(acetic acid) and 10.0mL of NaOH\n\n*This is a reaction so we have to use moles*\n\nHAc : 25.0mL * (0.100mol/L) * (1L/1000mL)= 0.00250 moles\n\nNaOH : 10.0mL * (0.125mol/L) * (1L/1000mL) = 0.00125 moles\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00125      0.00250        0\nChange:   -0.00125     -0.00125      +0.00125 \nEquilib:      0         0.00125       0.00125\n\nHAc and Ac-(NaAc breaks apart) for an equilibrium\n*Use Molarity when no reaction is taking place (best shown by when no volume is changing)*\n\nBoth HAc and Ac- have same molarity:\n\n(0.00125 moles / 35mL ) * (1000mL/1L) = 0.0357M HAc and Ac-\n\nUsing the equilibrium equation:\n\nEquation:  HAc   <--> Ac-     +  H+         Ka = 1.8 * 10^-5\nInitial:   0.0357M     0.0357M  \nChange:    -x          +x         x\nEquilib:   0.0357-x    0.0357+x    x\n\nKa=1.8*10^-5 = [Ac-][H+]/[HAc] = [0.00357+x][x]/[0.00357-x]\nKa=1.8*10^-5 = [0.00357][x]/[0.0357] usually good approx.\nKa = 1.8 * 10-5 = x = [H+]\n\npH = -log [H+] = -log [1.8*10^-5] = 4.745 (3 sig figs as only after the decimal counts for logs)\n\nB) 25.0mL HAc and 20.0mL NaOH\n\n*See A for HAc moles and double the NaOH for this trial*\n\n*This is a reaction so we have to use moles*\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00250      0.00250        0\nChange:   -0.00250     -0.00250      +0.00250 \nEquilib:      0            0          0.00250\n\nCalculate molarity of Ac-:\n\n(0.0250moles /45mL) * (1000mL/1L) = 0.0556M Ac-\n\nUsing the other equilibrium equation (Kb):\n\nEquation:  Ac- + H20   <--> HAc     +  OH-         Initial:   0.0556M   -        0\nChange:    -x        -        +x        + x\nEquilib:   0.0556-x  -       +x         + x\n\nThis gives us the Kb equation\nKb = ([HAc][OH-])/[Ac-] = with approx ([x][x])/[0.0556]\nso Kb = x^2 /0.0556\n\n**but we need to calculate Kb\n**Kw = 1*10^-14 at STP I'll assume this for this problem\n**Kw = KaKb so Kb = Kw/Ka\n**Kb = (1*10^-14)/(1.8*10^-5)= 5.56*10-10 \n\nso x = 5.56 * 10^-6 = [OH-]\n\npOH = -log [OH-] = 5.253\nPH = 14 - pOH (at STP)= 8.747\n\nC)25.0mL HAc and 30.0mL NaOH\n\n*See A for HAc moles and triple the NaOH for this trial*\n\n*This is a reaction so we have to use moles*\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00375      0.00250        0\nChange:   -0.00250     -0.00250      +0.00250 \nEquilib:   0.00125            0       0.00250\n\nAfter the reaction there is some NaOH left over so it is the only thing that matters for the pH as it is a stong base.\n\nCalculate NaOH molarity\n(0.00125moles/55.0mL)*(1000mL/1L)= 0.0227M NaOH\n\nStrong Bases dissociate 100% so [OH-] = 0.0227M\n\npOH= -log[0.0227] = 1.644\npH = 14-pOH = 12.356\n\nThis should show how they calculted the half-neutralization point (A), the neutralization point (B) and after the neutralization point (C).\n\nIf you have questions on this, feel free to send me a message.  I taught this a couple weeks ago so I'm still pretty fresh on the subject.
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.
512759
Do you mean M instead of N?\nIf you mean M, then\n1,00M/L*0,500 L= 0,500M\n0,500 moles NaOH*(40g NaOH/1 mol NaOH)= 20,0g of NaOH\nIt is c).
474593
Assuming 1 mole (28.01348 g) N2 = 22.4 liters at STP, I get (28.01348*.00220102/22.4) =  0.0027526 g if collected at STP. Correcting by 273/293 * (1.5*760-17.5)/760 (for temp., press. and water) = 0.003788 g. \nSo I basically agree with your results. What's the ans. supposed to be?
29056
Its' probably a rounding error. It would depend on which masses for the elements you use, how precise you want to be.
397103
the answer that i got is 3.10 M. \n\n3.25 g NaOH/40.00 g NaOH = .0813 mol NaOH\n\n26.2 mL / 1000mL = .0262 L\n\nM= .0813 mol / .0262 L = 3.10 M
515293
4.4.........the zeros cancel themselves otherwise the answer would be 8 but because of decimal point locations, its just 4.4.....\n\n\nhow did u come up with the answer?
361681
4.74 / 300 = 0.0158 g/ml\nin 35 ml\n0.0158 * 35 = 0.553 g\nDividing by 40 g/mol (MW) we get\n0.553 / 40 = 0.0138 mol in 100 ml\nSo in 1000 ml the number of moles will be 0.138\nThe molarity is then 0.138 M.
373673
C1V1=C2V2\n40%x100mL=25%xV2\n4000=25%xV2\n4000/25=V2   ((25%xV2)/25=V2)\n160=V2\nV2-V1=60\nshe should add 60 mL of water.
508169
800*0.15+x=(x+800)*0.32\n120+x=0.32x+256\nx-0.32x=256-120\n0.68x=136\nx=136/0.68\n=200\nhe has to add 200 ml
294565
No conversion is necessary.  36 ug/l equals 36 ng/ml, the ratio is the same!
470399
It won't matter how much you started with or how much higher the solution was heated.  When the solution cools to 20C it will hold 0.22g/100mL of solvent (water).\n\nSo if you start with 1.75g and 0.22g remain in solution, the amount crystallized at 20C will be the difference, 1.53g.
112458
too little\n\n0.25 liters = 250 milliliters\n\nyou need to add 50 milliliters more\n\nwhich 50 milliliters = 0.05 liters
55188
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
345637
FW of NaCl is 58.44  then\n\n   58.44 gm in 1000 mL = 1 Molar\n\n   4.78  gm in 200 mL = ?\n\ndo cross multiplication\n\n(4.78 gm x 1000 mL x 1 M) / (58.44 gm x 200 mL) = 0.4089 M\n\n(gm/gm and mL/mL cacels out so you get answer in M)\n\n\nNow you are taking 28 mL of 0.4089 M solution and diluting it to make final volume of 100.  Use M1V1 = M2V2\n\n0.4089 M x 28 mL = ? x 100 mL\n\n\nThen  ? = (0.4089 M x 28 mL) / 100 mL    { mL/mL cancels out and you get answer in Molarity)\n\n\n? = 0.1145 Molar is your answer
536982
0,220M/L*2,20L= 0,484 moles of HCl\n12,00M = 0,484 moles/?\n? = 0,0403 L or 40,3 mL of 12,00 M HCl solution + 2,16 L of pure water, so you can get you 2,20 L of 0,220 M HCl solution
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!