How can i determine the concentration of potassium carbonate?
87390
There are several possible answers to this question, depending on what resources you have available to you.  If the solution you are measuring is known to contain only potassium carbonate dissoved in water, you can measure the potassium concentration by either atomic emission or atomic absorption methods and calculate to concentration of the potassium carbonate.  You could measure the carbonate concentration by titration or colormetric methods and calculate the potassium carbonate concentration.  Or you could simply evaporate a known quantity of the water to dryness, weigh the residue and calculate the concentration from those numbers.  \n\nIf you have other compounds in the solution, the last option, gravimetric, pretty much disappears.  The first two options are still practical, unless the other compounds contain potassium or carbonate.  Then, things get a little more complex.
967763
Equivalent to [H+]^2/([HCH3CO2] - [H+]).  You can solve for [H+] knowing Ka and [HCH3CO2].
633715
5. KHCrO4 ----------------> K+   +   HCrO4-\nSince KHCrO4 dissociates complteley,\n the molarity of HCrO4- = molarity of KHCrO4\n the molarity of HCrO4- = 0.05M\n\nNow HCrO4 ------------------>    H+    +     (CrO4)2-\nCalculate the concentration of H+ from the pH value\n[H+] = 10^(-pH)   = 10^(-3.80)  = 1.58 * 10^(-4)\nIt is clear the only source of H+ is HCrO4-, therefore the concentration of (CrO4)2- is the same as concentration of H+\n[(CrO4)2-]   =   [H+]   = 1.58 * 10^-4\n\nKa = {H+} * [(CrO4)2-] / [HCrO4-]\nKa = [1.58 * 10^(-4)]^2/0.05\nKa = 5.02 * 10^ (-7)\n\n6.       KOH----------------> K+    +    OH-\n         HClO4 -------------> H+    +    ClO4-\n(a) \n    Moles of KOH = Molarity * Volume = 0.200M * 0.030L\n    Moles of KOH = 0.006 moles\n    From the equation above, \n    Moles of OH- = moles of KOH = 0.006 moles\n    \n    Moles of HClO4 = Molarity * Volume = 0.150M * 0.030L\n    Moles of HClO4 = 0.0045 moles\n    From the equation above,\n    Moles of H+ = moles of HClO4  = 0.0045 moles\n\n     H+    +    OH-    ------------>   H2O\nWhen this reaction is finished, 0.0015 moles of OH- are left\n[OH-]  =  moles of remaining OH- / Total Volume\n[OH-]  =  0.0015 moles/ 0.06L (30ml + 30ml)\n[OH-]  =  0.025M\npOH = -log [OH-]\npOH = -log (0.025) = 1.60\npH = 14 - pOH  = 14 - 1.60 = 12.40 \n\nDo the same for parts b, c, d, and e\n\n7. I will assume that \n   H2S --------------------------> 2H+    +    S2-\n   Ksp of HgS = 2.0 * 10^(-53)\n   Ksp of FeS = 6.3 * 10^(-18)\n\nKsp of HgS = [Hg2+] * [S2-]\n[Hg2+] = 0.01 M\n[S2-] = x\n2.0 * 10^(-53) = x*0.01\nx = 2.0 * 10^(-51)\nFrom the above equation, for every 1 mole of S2- there are 2 moles of H+\ntherefore,\n[H+] = 2 * 2.0*10^(-51) = 4*10^(-51)\npH = -log (4*10^(-51)) = 50.39\n\nKsp of FeS = [Fe2+] * [S2-]\n[Fe2+] = 0.01M\n[S2-] = x\n6.3* 10^(-18) = x* 0.01\nx = 6.3 * 10^(-18)\nFrom the above equation, for every 1 mole of S2- there are 2 moles of H+\ntherefore,\n[H+] = 2 * 6.3 * 10^(-18) = 1.26*10^(-17)\npH = -log (1.26*10^(-17)) = 16.90\n\nthe pH range is from 16.90-50.39\nThis is the best explanation I can gibe. By the way, pH level beyond 0 and 14 do exist, but, I'm not 100% about this answer.\nI fixed it, I think it is correct now.
573821
I think this question is incomplete because the Ka of HF hasn't been provided.  The amount of H+ that is formed from HF can only be calculated from its Ka value.
378344
At equilibrium, we have this below relationship between K and the concentration of all species.\n     \n　　　　　　　　　　　　　[H2]^2 [O2]\n　　　　　　　　　　　K= ----------------\n　　　　　　　　　　　　　　[H2O]^2\n\nSo we could get the answer by solving this equation.
729404
They yield K3PO4 and H2CO3.... Just switch the positive ion with the negative ion and make sure the charges cancel out to zero on each new salt.
851560
solution...Bicarbonate and sodium. :)
862382
Ah yes, the classic Iron Thiocyanate complexation lab.\n\nTo determine the concentrations for the first part, you need to know the concentrations of the stock solutions and the final volume of the solution you made.  You can then use the formula:\n\nC1*V1 = C2*V2\n\nWhere C is concentration and V is volume.\n\nFor the second part, I'm not really sure what you did in your lab.  Usually you use the calibration curve you constructed from the first part to measure the concentrations in the second part.  There is alot of ugly calculations to be done in there.  Check your lab packet for more information.
264737
Dissolve the following in 800ml distilled H2O.\n8g of NaCl\n0.2g of KCl\n1.44g of Na2HPO4\n0.24g of KH2PO4\nAdjust pH to 7.4 or 7.6 with HCl.\nAdjust volume to 1L with additional distilled H2O.
745984
To answer this question, you must know the Ka or Kb values for those acids or bases above.\nYou must then use those values to find the concentration of H+ (H30+ depending how you were taught it) or OH- ions.\nthese values can then be used to find pH or pOH depending which ion value you have.\nSimply perform the calculation \n\n-log[concentration H+ or OH-]\n\nthis will give you a pH value if you used H+ ions, or a pOH value if you used OH-. To get the pH from a pOH, you must take the value and subtract it from 14. (14-pOH = pH)
235402
See comment above.\n\nYou do need to do several trials at various concentrations though. You can then determine it through a linear regression.\n\nI also suggest that you consult a book on Analytical chemistry, because Beer's law has limitations. For example, it loses linearity at higher concentrations.
884595
For reaction aA <--> bB + cC, the Eq.constant K is defined as\n\nK = (B^b * C^c) / A^a\n\nSince in your equation a=b=c=1, and concentration for B=C, the equation simplifies to\n\nK = B^2 / A\n\nwith B=0.046 and A=0.454    (0.454=0.500-0.046)\n\nso K = 0.046^2 / 0.454 = 0.0047
1038474
http://www.chem.purdue.edu/gchelp/howtosolveit/Solutions/concentrations.html
639146
It will be the pH of the sodium hypochlorite formed.  Determine that amount, look up the Ka of the acid and solve.
830338
H2CO3 + 2KOH ===> K2CO3 + H2O\n2HNO3 + Ca(OH)2 ===> Ca(NO3)2 + 2H2O
803943
Number of moles of acid = volume (in litres) x molarity =0.0182 x 0.1  =  0.00182 Moles\nThe equation for the reaction shows twice as many moles of acid is needed as carbonate\n\n2HCl  +  Na2CO3  =  2NaCl +  H2O  + CO2\n\nSo the number of moles of Na2CO3 =  0.00182 divided by 2 = 0.00091\n\nThis is how many moles are in 10cm3 of solution so in a litre there would be 100 x as much  =  0.091 so the concentration is 0.091M\nConcentration is also measured in g/litre and in this case it is 106 (the RMM of sodium carbonate) x 0.091 = 9.646g/litre
433948
According to the MSDS for KCl,\nits solubility at 0 degrees C is 28.1 g/100 g of water