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[Comment by Poutnik] Important is also en.wikipedia.org/wiki/Grotthuss_mechanism for the proton interchange. Mobility of H3O+ and OH- gives a hint it must be fast. If you compare the diffusion coefficient of hydroxide ($\pu{5.270e9 m^2/s}$) to that of fluoride ($\pu{1.460e9 m^2/s}$), you might be surprised to see such a difference despite their comparable ...


5

I would guess $Ω^m$ notation is a personal invention and I couldn't find any sources that would standardize it. The choice of these symbols for abbreviating "equilibrium" may be justified that the glyph "Libra" ♎ originates from the Greek letter Omega Ω. Further, from Merriam–Webster Online: Equilibrium contains a root from the Latin libra, meaning "...


5

Since the equilibrium constant for the formation of the complex ion is very large, I assume that $\ce{[Ni(CN)4^{2-}] >> [Ni^{2+}]}$ From the comments: [comments:] That assumption is incorrect. The equilibrium constant seems large, but the exponents are high, so it is misleading. In fact, more of the cyanide is in the form of HCN than in complex ...


4

The Wikipedia article has a nice picture of the stages of deprotonation of phenolphthalein: https://en.wikipedia.org/wiki/Phenolphthalein However, there is a big jump here between H2In and In2-. In fact, smaller steps, showing just one deprotonation, are very reasonable. Whoever heard of two protons coming off at the very same time? http://www.ch.ic.ac.uk/...


3

I found this paper (it's in Japanese, but the relevant things are legible) https://www.jstage.jst.go.jp/article/yakushi1947/117/10-11/117_10-11_764/_pdf, which says that phenolphthalein has 2 pKas, pKa1 = 9.05 and pKa2 = 9.50. It also says that the pink form is the twice deporotonated; the 1- form exists, but is colourless.


2

AFAIK, such a reaction has no name. However, if one were so inclined, one could call it a replacement reaction, because the $\ce{MgCO3}$ is being converted into $\ce{Mg(OH)2}$. To calculate the "equilibrium point," which I take to mean the concentration of all the ions at equilibrium, we would need to know the $k_\text{sp}$ of both $\ce{MgCO3}$ and $\ce{Mg(...


2

Yes, the calcium ion could lead to precipitation. The solubility of $\ce{CaCO3}$ in distilled water is about 15 mg/L, which is about 0.15 mM calcium ion if there is no other source of carbonate. The solubility constant for $\ce{CaF2}$ is about $4\times 10^{-11}$, which means that we can only have 0.5 mM fluoride ions before precipitation will start. That's ...


1

I feel Koushal has not seen the difference between $\Delta G$ and $\Delta G°$, because $G$ of all reactants and products change during the reaction. $G$ and $G°$ are very different concepts. $G°$ is the Gibbs energy of $1$ mole of any reactant and of any product in the pure state at 25°C and 1 atm. $G°$ is a constant of the substance, independent of the ...


1

The Gibbs energy of reaction $\Delta_r G$ determines in which direction equilibrium lies, i.e. in which direction there has to be a net reaction (with a change in concentrations) to reach equilibrium. When equilibrium has been reached already, there is no net reaction (i.e. concentrations are constant). Nevertheless, at the molecular level, reactions in ...


1

If you look at the graph of the reactivity with relation to temperature it increases proportionally until it reaches its activation energy and then it proceeds to decrease. For exothermic reactions since the activation energy is met early the reactivity decreases when more temperature is provided. While for endothermic reactions since they require a higher ...


1

Consider three compositions: A. 2NaF + CaCO3 B. CaF2 + Na2CO3, and C. NaF + 0.5 CaCO3 + 0.5 CaF2 + 0.5 Na2CO3. Using data from the CRC Handbook (62nd ed), the heats of formation of A and B are respectively 560.47 and 560.6 kcal, so there is little driving force to make a reaction go to completion. Note that A should be near neutral pH, but B ...


1

The key word is "mixing". If you take a glass cylinder and fill it halfway with D2O (the heavier water, so put it on the bottom) and fill the cylinder ever so gently with H2O, the mixing will be determined by diffusion (NO mechanical mixing). But the experimental techniques required to analyze and determine the rates would be expensive and time-consuming. ...


1

Not really. It just depends on the reaction conditions. Consider the following reaction: $$\ce{N2O4 <=> 2NO2}$$ Here’s a picture showing this equilibrium in action: One of the reasons you can’t see the reactant here is because it is colourless, but this gives a pretty good example regarding visual equilibrium changes. this is a good answer giving ...


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The correct expressions are: $$\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2})\tag{1}$$ $$\ce{[H+]} = \sqrt{\frac{K_1K_2[\ce{HA-}] + K_1K_\mathrm{w}}{K_1 + [\ce{HA-}]}}\tag{2}$$ Equation 2 is an exact expression (neglecting activities vs. concentrations), but Expression 1 is an approximation. To dervice the approximate ...


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The former equation assumes $$[\ce{H2A}]\simeq [\ce{A^2-}]$$ due reaction $$\ce{ 2 HA- <=> H2A + A^2-}$$ The is possible with 2 simplifying conditions: The concentration of oxonium resp. hydroxide ions originated from water dissociation is much lower than concentration of the basic resp. acidic ampholyte form. $$[\ce{H2A}] \gg \sqrt{K_\mathrm{w}}$$...


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