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Sodium carbonate indeed dissociates in water : $$\ce{Na2CO3 <=> 2 Na+ + CO3^2-}$$ But that is just beginning, as in solutions with carbonates, bicarbonates or dissolved carbon dioxide happen multiple chemical equilibriums: ( See a lot of info at Carbonic acid on Wikipedia): Carbonate anion partially hydrolyzes in water: $$\ce{CO3^2- + H+ <=> ...


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The reaction system can be described as: $\ce{H2O <=> H+ + OH-}$ $\ce{H+ + CO3^{2-} <=> HCO3-}, \mathrm{p}K_\mathrm{a2}=10.33$ Note, the net of the first two reactions imply a rise in pH in the presence of carbonate. Further, with time and carbon dioxide exposure: $\ce{H2O + CO2 <=> HCO3- + H+}$ $\ce{H2CO3 <=> H2O + CO2}$ And, ...


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If one is already employing microwave (MW) assisted preparations, in this regard, per this 2007 source, 'Generation of hydroxyl radical in aqueous solution by microwave energy using activated carbon as catalyst and its potential in removal of persistent organic substances', by Xie Quan, et al, then one may, indeed, be able to increase the respective power of ...


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Active carbon is neither an acid, nor a base. It cannot. It does not react directly in water and in acidic or basic solutions. But it may have a catalytic action on the solutes. It may accelerate the rate of chemical reactions. That is probably what you have observed when mixing sodium bicarbonate in water. Without knowing it, you must have increased the ...


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The question asks: I need to create a $100\textrm{mL}$ buffer of $\textrm{pH = 9.20}$ with ammonia and ammonium chloride such that $\textrm{pH = }9.20\pm0.50$ with $20\textrm{mL}$ of $0.2\textrm{M } \ce{NaOH}/\ce{HCl}$. I am provided with $0.1\textrm{M}$ ammonia and ammonium chloride. User Mathew Mahindaratne has provided an answer, but did not answer ...


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Your initial calculations using Henderson-Hasselbalch equation is correct: $$ \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\ \mathrm{9.20} = 9.25 + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\ \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} = -0.05\\ \frac{[\ce{NH3}]}{[\ce{NH4+}]} = 0.89 $$ Yet, since you have only $\pu{0.10 M}$ ammonia ...


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Using concentrations and not activities, user GRSousaJr presents the exact solution to the problem in equations 1-8. However the simplifications don't really seem appropriate. This is a chemistry problem, not a problem in numerical analysis to solve. The gist is that we can make some simplifications based on the number of significant figures and a knowledge ...


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The chemistry is complex given the organic content of sea water and the influence of sunlight. For example, here is an article detailing the influence of sunlight on oceans containing dissolved organic matter, to quote: Solar radiation mineralizes dissolved organic matter (DOM) to dissolved inorganic carbon through photochemical reactions (DIC ...


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