What is the equilibrium equation for a buffer solution: of NaCO3& NaHCO3?
342463
HCO3(-) <---> H(+) + CO3(2-) (Sodium ions are spectators.)\n\npH can be determined by either using the normal Ka expression:  Ka = 5.6 x 10^-11 = [H(+)][CO3(2-)]/[HCO3(-)]\n\nor by the Henderson-Hasselbach equation:\npH = pKa + log [CO3(2-)]/[HCO3(-)]
711424
The Equation should be:\nNaCl + H2SO4 = Na2SO4 + HCl\nSo the coefficients are (2,1,1,2)
423379
let me see if i can at least help you with part of your problem, solubility in chemistry is simple, basically the equation will always be:\n\n[Solubility equalibrium constant] = concentration of product divided by concentration of reagents (concentration of each reagents multiplying by each other).\n\nthe easiest way to understand this is to simplify things for you so as you know, a chemical equation always go like this:\n\nreagents <==> products\n\nJust know this from the above, the dynamic here is that when you have an excess of either the reagents or the product, the balance of the equation tend to shift in the opposite direction (ie excess concentration of product will cause more reagents to form and vice versa), in a chemical equation such as the one from above, the reaction can actually go either way, products or reagents can form.  What really matters is the equalibrium constant that I talked about in the first equation. This constant represent the concentration of the products and the reagents when they are in equailibrium or when they are balance, meaning that the rate at which products or reagent form stays the same and there are no change in the concentration of either the products or the reagents so the whole thing is static at the point.   Think about this, more concentration of products wouldn't form unless you have extra reagents to spare that can shift the balance.  When something is saturated it means that it can longer dissolve in a solvent (let's use water to simplify the explanation)so it just accumulates at the bottom, water molecules can only separate so many ions from each other (the reagents) and preventing them from forming a solid product since water itself has limited molecules (if you mix enough table salt in a glass of water, it'll get to the point where it'll accumulate at the bottom and won't dissolve).  If remember correctly, in a chemical equation, when there's different amount of reagents needed to form the product, there's a slight modification to the equation I gave you, for example:\n\n2 CaCO3 + 4 ZnO2 <----> 3 ZnCO3 + 3 CaO2\n\nthen  [equilibrium constant] = [Cao2]^3 x [ZnCO3]^3 divided by [CaCO3]^2 x [ZnO2]^2\n\nnote that you should balance the chemical equations first, I am just giving an example.\n\nhere's a summary of how to compare numbers, if you plug given concentrations of reagents and product into the equation and the number you get is larger than the real equilibrium constant, then the concentration of products is higher than it should be and the equation will shift towards forming more reagents (in the case of the solid product, it'll dissolve) and if there's more reagents, more solid product is gonna form.  Hope that helps. Remember the Chemistry textbook is still your best reference.
367522
For pH3; Citrate buffer (Sodium citrate/citric acid) \n\nFor pH 8.75; Borate buffer (sodium tetraborate/boric acid), or TRIS [tris(hydroxymethyl)aminomethane]\n\nFor pH9.3; borate buffer, or carbonate buffer (carbonate/bicarbonate)
851560
solution...Bicarbonate and sodium. :)
853922
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)
410035
good news...10.63 is correct. Now how to get there.  \nMorp + H20 ---> Morp+ + OH-\ninitial concentration of Morp=.115M  everything else is 0.\nEquilibrium concentration of Morp=.115M-X, H20 ignore, Morp+=X, OH-=X.   Set up your equilibrium equation.  Kb=[OH-][Morp+]/[Morp]  pKb=-log Kb  so Kb=1.62e-6.  1.62e-6=[X][X]/[.115-X]  We can ignore the X subtracted from .115 since it is very small when compared.  so solving for x we finally get X=4.32e-4 which is the OH- concentration.  pOH=-log OH-  pOH = 3.36.  pH = 14-pOH .....pH=10.63.
205870
CaCO3(s) + 2HC2H3O2(aq)-> Ca(C2H3O2)2(aq) + H2O(l) + CO2(g)
844168
This is very simple\n\nCa+H2O -> Ca(OH)2 + H2\n\nThe reactant has\n1 atom of Ca\n1 atom of H\n2 atoms of O\n\nThe product has\n1 atom of Ca, ok\n2 atoms of H\n2 atoms of O\n\nJust put a coefficient of 2 in the water (H2O)\n\nSo the chem. eqn becomes\n\nCa+2 H2O -> Ca(OH)2 + H2\n\nThat's it.\nThe equation is already balance
304632
Okay, so the first thing that happens it that the NaF completely dissociates:\nNaF ---> Na+ + F-\n\nCool.  Now, the Na+ is just floating around, so we can ignore it.  Let's deal with the F-.  The F- is in the water, so it's going to do this:\nF- + H2O <---> HF + OH-\nNote that the arrow is going in both directions; it's an equilibrium problem, so I need to set up an equilibrium equation.  \n\nBut first, because I'm dealing with OH- ions, I'll need a Kb value.  Since Kw, the product constant of water, equals Kb * Ka, and Kw= 1E-14:\n(1E-14)/(6.8E-4)= Kb = 1.47E-11\n\nNow,\nKb=[OH-][HF]/[F-]\nor\n1.47E-11 = x^2/(.00339-x)\nsolve for x (I just set it equal to zero, graph it, and find the x-intercpets on my calculator):\nx=2.23E-7 M \nSince x=[OH-], [OH-]=2.23E-7 M\n\nWe can calculate pOH by taking the negative log of [OH-]:\n-log 2.23E-7 M = 6.65\nSince pH + pOH = 14.00:\npH + 6.65 = 14.00\npH = 14.00-6.65\npH = 7.35, the answer within one one-hundredth, which is probably some minor rounding discrepancy.  Yay!
433273
Na2CO3 + 2HCl = CO2 + H20 + 2NaCl (aq)\n\nThis reaction is a double displacement reaction because oxygen from the sodium carbonate combines with the hydrogen in the hydrochloric acid to form water, and the sodium goes with chlorine to form sodium chloride, making carbon dioxide in the process.
600985
According to the Henderson-Hasselbalch equation,\n\npH = pKa + log [A-]/[HA]\n\nSo, when pH = pKa, then [A-] = [HA]\n\nA buffer has is a mixture of undissociated acid and the products of dissocation.  With the concentrations equal, there is the maximum number of ions to react with to changes in pH.
1029892
Na2S (aq) + 2H2O (l) = 2NaOH + H2S (g)
474059
Gold metal won't react normally with sodium carbonate at all.  \nIf you're talking about a gold ion, though:\nAu+ + CO3(2-) ---> Au2CO3\nThe sodium is always soluble, so it is ignored in net ionic equations, which makes the first answer to this question incorrect.
1097630
Na2CO3 + 2HCl-> NaCl + H2O + CO2
1136961
Are you preparing a buffer soln with NaHCO3 and NaCl ? Then take equimolar amounts of both and dissolve them in 100 ml of water and then dilute them to the desired strength.
780228
HCL and NaOH are 'strong' acid and base. When they react, the product is neutral. NaOH + HCL --> NaCl + H2O\n\nWhat you need to find is which is the limiting reactant, the other one (what is left over) will effect the pH by its concentration in the otherwise neutral solution.\n\nFor your first problem:\n\n0.0557M HCL = 0.0557mols/liter*0.00542L = 3.02*10^(-4)mol HCL\n\n0.0377M NaOH = 0.0377mols/liter*0.00337L = 1.27*10^(-4)mol NaOH\n\n3.02*10^(-4)mol HCL - 1.27*10^(-4)mol NaOH = 1.75*10^(-4)mols HCL\n\n0.00542L + 0.00337L = .00879L\n\n[(1.75*10^(-4)mols HCL) * (1 L) ] / (.00879L) = .020M HCL\n\npH = -log(.02) = 1.70