1.0 M NaH2PO4 and 1.0 M Na2HPO4 are in a 10.0 L container. the pH is 7.21. What is the pH if .50 mol NaOH is added, assuming no change in volume?
204107
Well first of all you have not given us all the information to be able to answer this buffer question.  We need to know the Ka (6.2x10-8)of the acid.  But the steps to perform the task.  Calculate how much of the acid is neutralized by the addition of the base.  Recalculate the equilibrium concentration of both the acid and base with the new acid value.  using Ka=[H+][Na2HPO4]/[NaH2PO4] solve for H+ and finally use pH=-log[H+].  Your final answer..pH=7.25
495164
step 1 2NaOH + H2SO4 ---------> Na2SO4 + 2H2O\nstep 2 amount of NaOH = 0.2 x 25/1000= 5.0 x 10^-3 mol\nstep 3 amount of H2SO4 = s.0x10^-3 x 1/2 = 2.5 x 10^-3 mol\nstep 4 mass of H2SO4 = 2.5 x 10^-3 x 98 = 0.245g required\n\nNOTE\n^= power\nin step 4 it is multiplied by 98 because that is the molar mass of H2SO4 ie (2x1)+32+ (16 x 4) =98\n\ni do hope this is correct it has a been about 3 yrs since i did my A Level in chemistry
494559
It seems that the tripolyphosphate hydrolises to the orthophosphate\n\nThis is the only way I could see it balance BUT I am guessing and balancing ...\n\n3Na5P3O10 + 6H20 → 5Na3PO4 + 4H3PO4
625024
Your question is not very clear, however I am attempting it.\n1:  1 mol of Na3PO4 will have 3 mol of Na and 1 mol of PO4 ions.\n2:  2.5 mol of Na2SO4 will have 5 mol of Na and 2.5 mol of SO4 ions.\n3: 3 mol of CaSO4 will have 3 mol of Ca and 3 mol of SO4 ions.\n4: 4 mol of NH4NO3 will have 4 mol of NH4 and 4 mol of NO3 ions.
391866
d. \n\nIt's a double replacement (acid/base) reaction.  The Sr and the H are both positive and they switch.  Since Sr has a 2+ charge and the PO4 has a 3- charge, you use a common multiple of 6.  The other product would be H(OH) or water.
226263
P3O10 does not exist; it's P4O10.\n\nBalanced equation is:\n\nP4O10 + 6H2O --->  4H3PO4
338927
IAnd for your other request, I would request you to send a seperate mail to me through yahoo Answer so that I am able to send it part by part.These are long subjects and would take some time,The balanced equation would be                                           H3PO4+3KOH=K3PO4+3H2O
251519
U ripped it of from da AISSE 2006 sample que paper didn't ya\na) pH = —[H+] ½\n= — log (1 x 10—10)\n= — (—10)\n= 10 ½\nb) basic ½\nc) decrease ½\nSince the concentration of H+, [H+] increases, — log10 [H+] decreases,\nso pH decreases
516097
to what?  Are you looking for the molecular weight of the monoprotic acid?  If so, remember M (molarity) is in moles/lter) and your colume of NaOH is in mL so you'ww want to convert that to liters.\n\n\n.01524 liters x .1578 moles/liter = 0.002405 moles\n\nso therefore you have 3.567  g / 0.002405 moles = 1483 gram/mole\n\nuse dimensional analysis and keep track pf the units to find the answer
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
321686
NH4ClO4 --> NH4^+ + ClO4^-\n\nNow in a water solution,\nClO4^- + H2O --> HClO4 + OH- --> This doesn't happen because HClO4 is a strong acid.  Therefore the only contribution to the pH is from the NH4^+\n\nNH4^+ <=> NH3 + H^+\nYou start with [NH4^+] = 0.40 M, and [NH3]=[H^+] = 0 M; after undergoing equilibrium [NH4^+] = 0.40 - x, while [NH3]=[H^+] = x; Ka = [NH3][H^+]/[NH4^+] = (x)(x)/(0.40-x) = Kw/Kb = (1 x 10^-14)/(1.8 x 10^5)\nAssume x << 0.40, so x^2/0.40 = 5.56 x 10^-10\nx = 1.49 x 10^-5 M = [H^+]\npH = -log(1.40 x 10^-5) = 4.827
376020
H3PO4 + 3NaOH ----> 3H2O + Na3PO4
318012
NaOH + HCl --> NaCl + H20\n\nn(NaOH) = 1.00 x 32.6cm3\n        = 1.00 x 32.6 x 10^-6\n        = 3.26 x 10^-5  mol\n\nn(HCl) = 1.00 x 25cm3\n       = 1.00 x 25 x 10^-6\n       = 2.5 x 10^-5  mol\n\ntherefore, NaOH is in excess.. hence the final solution is alkaline, which means the pH is more than 7.\n\nDue to the limitation of my knowledge, i can't give you the exact pH value. Just hope that it will help you to continue solving...
228046
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
598186
I think this is how you do it:\nacid and base is measured in terms of pH and pOH. as you know, the H makes it acidic and the OH makes it basic, and if you mix an H and OH, you get water, which is to say it was neutralized.\nSo, let's say you have HBr with a pH of 4. You that pH is a measure of the concentration of H present in the mixture. So, to get the actual number you do -10ph, and you'll get a number like 10^-5 or something. So, to neutralize that many Hs, that's how many OH's you need. so, you'll need a pOH 4, which is the same thing is a pH of 10 (pOH = 14 - pH). \nhope that helps, good luck. and read the book, and do some of hte exercised in the book.
354191
First off, you need the concentration of the NaOH solution.  From that you can determine the number of moles of NaOH that were used.  This is equal to volume (V) time the concentration, in mol/L (C).\nn = V * C = 0.02260 L * ? mol/L = ? mol\nn = 0.02200 L * ? mol/L = ? mol\n\nThe reaction of KHP (C8H5O4K) and NaOH is this:\nHOOC-C6H4-COOK + NaOH -> NaOOC-C6H4-COOK + H2O\nSo, 1 mol of NaOH will react with 1 mol of KHP.\nThat means that the number of moles of NaOH you calculated above is also equal to the number of moles of KHP.\nThe molecular weight of KHP is 12*8 + 1*5 + 16*4 + 39 = 204 g/mol\nMultiply that by the number of moles: 204 g/mol * ? mol = g actual KHP\n\nYou can divide that number by the 0.911 and 0.950 g and then multiply by 100% to find the purity.\n\nHere’s an example if your NaOH solution was 0.1 M (M means ‘molar,’ or mol/L):\nn = 0.02260 L * 0.1M = 0.002260 mol of NaOH = 0.002260 mol of KHP\n204 g/mol * 0.002260 mol of KHP = 0.461 g pure KHP\n0.461 g pure KHP / 0.911 g impure KHP * 100% = 50.6 % KHP\n\nHope this helps!
259381
Well to determine [H+] you use the equation that \npH=-log [H+] \nthus a solution with pH=3 will have a [H+] of 1x10^-3, \nand a solution with pH=8 will have a [H+] of 1x10^-8\nTo find [OH-] there are several equations you can use\n[OH-]=Kw/[H+], \nKw is the ionization constant of water which is equal to 1.0x10^-14\nThus you will have [OH-] of 1.0x10^-11 and 1.0x10^-6\nSo you answer will be that the solution with pH=3 will have a [H+]>[OH-]