What is the Ka of the acid?
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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
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Potassium Hydroxide and Hydrochloric acid.\n\nKOH + HCl = KCl + H2O
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It's a salt - ammonium acetate. Ammonium is an acid, but the acetate ion is actually alkaline.
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below 7 is acidic\nabout 7 is neutral\nabove 7 is basic\n\nthe scale ranges from 0 - 14
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<--------- 7 ---------->   (pH)\nacid                base\n\nthe smaller pH the stronger acid\nthe bigger pH the stronger base\n\nso you should arrange them from the smallest pH to the biggest\nstomach content (1.6)\nsoda pop (3)\nmilk (6.5)\nfresh water (7, of course neutral)\nhuman blood (7.4)\nsea water (8.7)\ndrain cleaner (8.7)
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Look up the chemical in a pKa ( -log of dissociation constant K)table of acids. H2PO4- has a pKa of 7.20. It is not very acidic.
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Dilute acetic acid (CH3COOH) is calle as vinegar (I think 30% dilution)\nSince acetic acid is weak acid, its pH will be around 3.5 to 5\n\ndefinately its not 2.4.... \n\nYou can say for vinegar, pH is around 4 to 5
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Chemicals used in the reaction is important. What is the pKa of the reactant? Research on which buffer and what its pH buffering region is important, because it should cover the pKa region of the chemicals used. Once knowing which buffer to make, we can take a buffer by making a mixture that have a weak acid with its salt of conjugate base, or weak base with its conjugate acid. It should be noted that here buffer solutions are presented in terms of the Brønsted-Lowry notion of acids and bases.
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pH = pKa - log (.25/.4) ~= 4.3
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Acid = Hudrochloric acid (HCl), Sulfuric acid (H2SO4)\n\nBase = Sodium Hudroxide (NaOh), Potassium Hydroxide (KOH)
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Are you sure this is a weak acid? Because to solve a weak acid dissociation problem you'll need the Ka value for that specific acid. Strong acids dissociate fully, so those are easier.\n\nFor a strong acid, it's simple stoichiometry. The HY acid dissociates completely, and you'll end up with 0.02 M HY left.\n\nFor a weak acid, you'll need to set up an ice table, with the K value.\n\n   HY + OH- --> Y- + H2O K = ?\nI 1.0  0.980    0    ---\nC -x   -x       +x   ---\nE 1-x  .98-x    x    ---\n\nSo, the equation you'll get is K = x/((1-x)(0.98-x)). If you can find the Ka value (from a book or something, then you can figure out the rest of the values by solving the equation.\n\nEDIT: My table isn't displaying correctly, but just move those values over to the correct columns.
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What are the pka's of all the different species? It sounds like the end product is a mess of conjugate bases, so the pka's are needed to figure out which disassociation reaction will predominate.
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These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
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first, you find out the equation of Keq and delta G. I think it should be Keq = e^ [-deltaG/(RT)].\nThis equation and constant R can be found in any general chemistry textbook. And T is usu. 25+273=298K\nthen you can find out deltaG for the equation.\n\nwe also have deltaG= [- delta G for Ag(NH3)2^+1(aq) + (-32.0)] - [-26.2 - 3.8]\n\nand there comes the answer!!! ;)
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2KOH + Cl2 ----> KCl + KClO + H2O\n6KOH + 3Cl2 ----> 5KCl + KClO3 + 3H2O
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Look at their conjugate bases and whichever is more stable will have the weakest acid. PKA values are also easy to prove this becuase it is a realtion of the H+ content.
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Kp=Kc(RT)n\na.Kp=Kc(RT)-1/2\nb.Kp=KcRT\nc.Kp=KcRT\nd.Kp=Kc(RT)2