what is the pH of a solution that is .50 M in sodium acetate and .75 M in acetic acid?
440505
pH = pKa + log ([Salt]/[Acid])\n\n[Salt] = 0.5 ; [Acid] = 0.75 ; ([Salt]/[Acid]) = 2/3\n\npKa = -log (1.85 x 10^-5) = 5-0.2671\n\nlog ([Salt]/[Acid]) = log 2/3 = -0.1761\n\nHence, pH = 5 - 0.2671 + (-0.1761)\n          = 5 - 0.4432\n          = 4.5568\n\nNote: All Calculations done using Clark's Tables. Use your calculator to calculate log for greater accuracy.
819244
a) Acid +H2O -> H3O(+)+ Basis\n\nKa1 =  x²/(0.1-x) = 10-4\nx<< 0.1 so x = 10-3\n\nso pH = 3\n\n2) second dissociation\nKa2 = y (10^-3 + y) / (10^-3 -y)\ny << 10¨-3\ny = Ka2 = 10^-8
741771
0.5M = 500mM\nv x c=v'xc'\nv x 500 = 50 x 1\n0.1mL of the 0.5M solution diluted to 50mL will be 1mM
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.
914456
okay...100 g*(mol/12.01g)=8.33 mol C\n\ntake the aforementioned 1mol *(-113kj/2mol)=-56.5 kJ released for 1 mol\n\nthen multiply that -56.5 by 8.33= -470.65 kJ released\nnote it is exothermic and thus negative
1171925
PH of the aqueous solution of NaCl is 6 - 7 @ 1% by weight
608403
NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)
822325
a) Find some Acetic Acid (CH3COOH)\nb) Find some Copper (II) Hydroxide (Cu(OH)2)\nc) Add to form Cu(CH3CO2)2 + 2H2O\n\nWithout the volume or the number of moles, preparation of a 0.530M can be done several ways. \n\n2CH3COOH + Cu(OH)2 --> Cu(CH3CO2)2 + 2H2O\n\nUsing 1L:\n0.530 mol of Cu(CH3CO2)2 means 1.060 mol of CH3COOH\n\n1.060 / 0.5L = 2.120M\n\n0.530 mol of Cu(CH3CO2)2 means 0.530 mol of Cu(OH)2\n\n0.530 / 0.5L = 1.060M\n\nSo mix 500mL of 2.120M Acetic Acid with 500mL of 1.060M Copper (II) Hydroxide.\n\nN.B. Being a weak acid, acetic acid will not fully react with copper hydroxide, and hence the actual molarity requiredmay be higher.
490327
solution A has a hydrogen ion concentration of 10^-5\nsolution B has a hydrogen ion concentration of 10^-3\nAll you have to do is 10^-ph\n\nIs this homework?
169137
We have 48 gallons of a solution 64% acid\nSo the acid quantity = 48*0.64\n\nWe will add X gallons of a solution 16% acid\nSo the acid quantity = X*0.16\n\nThe Total acid quantity= 48*0.64 + X*.016\nThe Total Volume = 48 gallons + X gallons\n \nThe Average Percantage of acid= Total acid quantity/total volume\n=[48*0.64 + X*0.16]/[48+X] ===== 0.52 as given\n\n48*0.64 + X*0.16 = 0.52*(48+X)\n48*(0.64 - 0.52) = (0.52 - 0.16)*X\n48*0.12 = 0.36*X\nX=48 * 0.12/0.36= 48/3 = 16 gallons\n\nWe must add 16 gallons of solution 16% acid
774152
63.33 mL required.\n\nTo prepare the solution, you need 0.380 M * 2.00 L = 0.760 moles sulfuric acid.  To obtain 0.760 moles of sulfuric acid from a 12M sulfuric acid solution, you need 0.06333 liters, or 63.33 mL.
710157
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
1029892
Na2S (aq) + 2H2O (l) = 2NaOH + H2S (g)
255966
Use the hendeson-hasselbach equation\n\npH = pKa + log([A-]/[HA])\n\nand that you know the toatal concentration of ions\n.2 = [HA] + [A-]\n\nyou know the pKa (found in textbook or given) 4.18\nthe pH is 4.00\n\nso -.19 = log(A-/HA)  therefore .6457 = A-/HA\nand .2 = [HA] = [A-]\n\n... solve simultaneous equation\n\n[HA] = .122 and [A-] = .078\n\ndouble check\n\nK = ([H+][A-])/[HA]\n\nK = 10^-pKa = 6.45E-5\n[H+] = 10^-pH = 1E-4\n...and it does check out
906606
1) To prepare this solution you should do:\n\n100 g of solution needs 2.5 g of solute (to get 2.5% mass/mass)\n150 g of solution needs  x    g of solute (to get 2.5% mass/mass)\n\n2)  x = (150 g x 2.5 g / 100 g) = 3.75 grames of KCl\n\nThat's it!\n\nGood luck!
586833
the vinegar is a solution of water and acetic acid.  the acetic acid exists as acetate ion and hydrogen ion.  the bicarbonate is a sodium ion and a bicarbonate ion.  when the two combine, the postive hydrogen ion combines with the negative bicarbonate ion to produce carbon dioxide and water, and the negative acetate ion and the postive sodium ion combine (well, they really don't combine, they stay apart and disolved in the solution) to become sodium acetate.  what's left is a water solution with sodium acetate desolved in it and bubbles of carbon dioxide coming out of the solution.
769229
At the equivalence point the curve changes its curvature. Thus instead of trying to find the middle of the curve in order to determine the endpoint you can find it by doing the 1st derivative (the endpoint will be a peak) or the 2nd (at the endpoint it will be 0))\nSince you are using experimental data, when you are calculating the derivatives you will have an error depending on the amount of base added at each step (the bigger the volumes, the bigger the error; remember that detivatives require by definition extremely small changes in order to be accurate). I guess your instructor told you to take the average of the two in order to minimize the error.\n\nThe equation is actually pH=pKa+log[conj.base]/[acid]. At half equlivalence point  [conj.base]/[acid]=1 thus pH=pKa+log1=pKa.\n\nIf your equivalence point is at 6.97 ml then the half equivalence point is at 3.48-3.49 ml. Go to your pH vs V base plot, and find the pH value for that volume. For that value pH=pKa and thus Ka=10^-pKa