which of the following a Buffer solution?a.HCL and KClb.NaOH and NaClc.HNO3 and NH4NO3?
373054
HNO3 and NH4NO3. A buffer is a solution of a weak acid with the corresponding weak base. It is set up to maintain a specific range os Ph, determined by the proportions of acidvs base and by the POh of the acid/base. The amoutnt of H+ that a buffer syatem can absorb before the buffer breaks is determined by the streangth, ie molarity, of the solution. a
664234
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
633022
NaCN-Sodium Cyanide-alkaline\nNH4NO3-Ammonium nitrate-acidic\nNH4Cl-Ammonium chloride-basic
696123
buffer solution ?\nBuffer solution is  solution of which the pH does not change significantly on the addition of a little acid or base.\n\nBuffer are also important in many industrial processes.Certain processed food has to be prepared in a buffered form so that when consumed does not cuase undue changes in pH.
747027
It'd be basic speaking qualitatively. NH3 is a base so any amount in water would make the solution basic.
350928
AMMONIUM ion is NH4+, people!  Check your books occasionally ey lads?\n\n(NH4)2 CO3 -->  2 NH4+   +    CO3 2-\n\nNH4+ is more than 10,000 times weaker an acid than acetic acid; CO3 2-  will form HCO3-  so i'd say the solution will be basic ( but only weakly! )
426896
BaCO_3+HCl - BaCl_2+CO_2+H_2_O\n\nbalance it
419911
That's easy. [Co(NH3)6]Cl3. This is coordination chemistry, dude. There can be other products depending on the ammonia concentration. But this is your best bet. I've been doing these reaction in the lab way way back.
734957
Na+ Cl-\nC +4 O-2 each\nK+ Br-\nN-3 H+1 each\nMg+2 O-2
1051035
1-. NaCl + H2O Sodium Chloride ( table salt )\n2-. NaSO3 +2 H2O Sodium Sulfate ( or sulfite I always forget which is the 3 and which is the 4 )\n3-. NaC2H3O2 + H2O ( I'm guessing.  You will want to check your periodic table for the charges on this one )\n4-.NH4Cl + H2O.  Ammonium Chloride\n5-.NSO2 + 3H2O\n6-. not sure\n\nI presume you are working salt solutions.  Ammonia was never my strong suit. I'm positive about the first two.
172666
When a salt derived from a strong acid and a weak base dissolves in water, the solution becomes acidic. For example, consider the process\n \nNH4Cl(s) ----H2O ---> NH4+(aq) + Cl-(aq)\n\nThe Cl- ion has no affinity for H+. The ammonium ion NH4+ is  a weak conjugate acid of the base NH4+(aq) and ionizes as follows \n\nNH4+(aq) + H2O(l) <=> NH3(aq) + H3O+(aq)\n\nor simply  \n\nNH4+(aq) <=> NH3(aq)  + H+(aq)\n\nThe equilibrium constant ( or ionization constant) for this process is given by \n\n\n\nKa = Kw / kb  =   1.0*10^-14 / 1.8*10^-5 = 5.6*10^-10\n\n\n      x^2 / .10 =  5.6*10^-10\n\n           x^2  =   5.6*10^-11\n\n             x  =    7.5*10^-6 M  =  [H+]\n\npH  =  -log [H+]\n    \n    =  5.12\n\n     my answer is  C , NH4Cl
838903
This is just an acid-base reaction.  The nitrogen lone pair acts as a Lewis base and takes the HCl proton.  You then have a quaternary ammonium cation and a chloride anion.  This will form the salt shown as answer b.
883220
H2+O2-->H2O\nNaOH+HCl-->NaCl+H2O\nKOH-->(on heating,deomposition)K2O+H2O\ni guess this is what u want\nand i f u didn't get it i am sorry......
1168891
The balanced solution is nothing else than a special physiological salt solution.  Near the sodium chloride it generally contains CaCl2, KCl, potassium and sodium-phosphate, Mg-salt(chloride or sulphate), sodium hidrocarbonate.\nAs additional component it may contains some glucose and/or phenol red as a pH indicator.
836404
No, plain water is not adequate.  It has no buffering potential.  A buffer is a acid or base dissolved with a sufficient quantity of its salt to accept significant additions of acid or base while resisting drastic pH change.\n\nThere are lots of tables you can find in most inorganic chemistry textbooks, technician's handbooks, the Handbook of Chemistry and Physics, the Merck Index, etc that tell you what solutions to make and how to mix them to achieve a buffer with good activity at any given pH.  I don't have a library in my room and I can't really find a good websource so you'll have to do the legwork.  Prepare the solutions and combine them as stated.\n\nThen, based on the components, you will have to determine if the molarity (of what?) matches the ionic strength you have.  I have provided a link to a webpage that describes the ionic strength calculation.\n\nGood luck.
621533
NaOH ----------------> Na+   +   OH-\nFind the number of moles of NaOH\nMoles of NaOH = 0.10M * 0.150L = 0.015 moles\nSince the coefficients in fornt of NaOH, Na+ and OH- are all 1, therefore if 0.015 moles of NaOH disssociates, 0.015 moles of Na+ and 0.015 moles of OH- are produced.\n\nSimilarly we will work with HCl\nHCl --------------------> H+      +      Cl-\nMoles of HCL = 0.30M * 0.05L = 0.015 moles\nThis means the 0.015 moles of H+ will be produced\n\nSince H+ and OH- have equal amiunts, they will completely neutralize each other leaving the solution at neutral pH. Since the neutral pH is 7, therefore, the pH of this solution will be 7.
160963
K2SO4 ---> 2K+  +  SO4-\n\nHCl ---> H+  +  Cl-\n\nNaHCO3 ---> Na+  +  HCO3-\n\nNH4OH ---> NH4+  +  OH-\n\nBaCl2 ---> Ba-2  +  2CL-\n\n(NH4)3PO4 ---> 3NH4+  +  PO4-3