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If you dissolve $\ce{NH4Cl}$ in water, your solution contains $\ce{NH4+}$ and $\ce{Cl-}$ ions. If you dissolve $\ce{NaOH}$ in water, your solution contains $\ce{Na+}$ and $\ce{OH-}$ ions.
Now if you mix those two solutions, nearly all $\ce{NH4+}$ and $\ce{OH-}$ will react and produce $\ce{NH3}$ and $\ce{H2O}.$ This shows that $\ce{NH4OH}$ cannot exist. It is ...
5
You are talking about two different things and phenomena. One is leaching-which means chemicals in the plastic dissolve into the solution with the passage of time. The other thing is literally physical pieces of microplastics- they are not dissolved but they are actual microscopic pieces of plastic just like very fine dust is present in air. It is not ...
5
As orthocresol suggested, to illustrate "classic" acidity as it's understood on a school level you unquestionably want to start with Arrhenius definition, which attributes acidic properties to any substance which increase concentration of hydronium $\ce{H3O+}$ ions in water.
That's said, the very first reaction you probably want to write is dissociation of ...
4
There are 3 equilibriums, described by 3 equilibrium constants:
The first one is the equilibrium of water dissociation:
$$\ce{H2O <<=>H+ + OH-}$$
$$K_\mathrm{w}=[\ce{H+}][\ce{OH-}]$$
The second one is for the ( very weak ) hydrogencyanide acid dissociation:
$$\ce{HCN <<=>H+ + CN-}$$
$$K_\mathrm{a}=\frac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]...
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I am not sure what back bonding you are referring to. It may be related to the outdated concept of silicon using its vacant, high-energy d orbitals in any meaningful way which is not actually the case.
There is a second effect which consists of overlap of nitrogen’s p orbitals with the σ* orbitals of the $\ce{Si-H}$ bonds (which are silicon-centred ...
4
"Microplastic" breaks off from solid plastic material by mechanical wear, often following a degradation of the material by environmental influence like oxygen, light, ...
The inside of reusable water bottles does not experience mechanical wear. Single use bottles even less. Any microplastic particles in there must obviously have come with whatever beverage ...
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Op's question is what is the correct $\mathrm{p}K_\mathrm{a}\ce{H}$ value of pyrrole. Thus, I'm not going to elaborate OP's findings, but would try to give a reasonable answer to the question. The most reasonable answer I found for $\mathrm{p}K_\mathrm{a}\ce{H}$ value for the pyrrole is $0.4$ (for the novices, this is the $\mathrm{p}K_\mathrm{a}$ of the ...
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To quote the key points of an easy to read publication by Salgado and Vargas-Hernández (doi 10.4236/ajac.2014.517135, open access):
All starts with the dissociation equilibrium between the acid $\ce{HA}$ and its anion $\ce{A-}$ for which you write
$$\ce{HA + H2O <=> H3O+ + A-}$$
By consequence of this, the recorded total absorbance $A_\mathrm{t}$ ...
3
Both -$\ce{OCH3}$ and -$\ce{OH}$ groups have exhibited two effects on the aromatic ring: (1) Electron donating resonance or mesomeric effect (+M) and (2) Electron withdrawing inductive effect (-I). For both acids in hand, the electron withdrawing inductive effect (-I) is almost same since both -$\ce{OCH3}$ and -$\ce{OH}$ groups are 4 carbons away from the ...
3
In this case, cerium(IV) sulfate is a strong oxidising agent, hence, using $\ce{HCl}$ would likely be oxidised to form $\ce{Cl}$ of higher oxidation state. Therefore, the titre would be vastly higher than expected.
$\ce{HNO3},$ on the other hand, is an oxidising agent and would oxidise $\ce{Fe}$ to a higher oxidation state than expected (of $\ce{Fe^3+}$). ...
3
$\ce{H2A}$ and $\ce{A^2-}$ forms of a biprotic acid seldom coexist as major components, expecially if the acid is weak, unless $\mathrm{p}K_\mathrm{a1}$ and $\mathrm{p}K_\mathrm{a2}$ are similar. Instead, coexistence of major components $\ce{H2A / HA-}$ or $\ce{HA- / A^2-}$ occurs.
$\ce{SO3^2-}$ starts to form in about neutral or alkalic solutions, while $\...
3
It is not a chemical problem, but a trivial mathematical problem.
If you have an equation $x \cdot y = c$, where $x$, $y$ are variables and $c$ is constant, then if $x$ increases, $y$ must decrease, otherwise $c$ is not a constant.
The key part is to understand that chemical equilibrium means existence of 2 ongoing opposite chemical reactions of the same ...
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Yes. If some acid is added to pure water, $[\ce{H+}]$ increases but $[\ce{OH-}]$ decreases. It means that:
in pure water, enough $\ce{H2O}$ will break into $\ce{H+}$ and $\ce{OH-}$.
in the presence of an acid, a smaller amount of water will break into $\ce{H+}$ and $\ce{OH-}$. The presence of an acid prevents $\ce{H2O}$ from being dissociated.
For ...
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Please be aware that adding salt changes the ionic strength of your solution. Therefore, the activity $a_\mathrm H$ of $\ce{H+}$ is changed. This changes $\mathrm{pH}=-\log_{10}(a_\mathrm H)$. In a 1 molar $\ce{NaCl}$ solution ($I\approx 1\ \mathrm{mol/l}$) the Debye-Hueckel (DH) approximation gives a shift of the activity by around 0.5 pH units (compared to ...
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Not according to the commonly accepted mechanism here
The first step is
The nucleophilic H from the hydride reagent adds to the electrophilic C in the polar carbonyl group of the ester.
Note: the pKa of primary amides is in the 22-25 range so not close to alcohols.
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I am directly going to address the source of your confusion: $\text{+M}$ of -$\ce{OCH3}$ is more than -$\ce{OH}$. You see, you are basically saying that the $\text{+I}$ effect of $\ce{CH3}$ in -$\ce{OCH3}$ is going to push more electron density towards the ring, hence, -$\ce{OCH3}$ is more activating than -$\ce{OH}$.
But look at the bigger picture, and let'...
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You may pour them down the sink. NaOH will be soon neutralized and destroyed by the CO2 from the atmosphere, or by the bicarbonate ions present in all drinkable calcareous waters.
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If you're dissolving into neutral water, then you are overthinking the problem.
If you dissolve a weak acid into neutral water, the result is an acidic solution.
If you dissolve a weak base into neutral water, the result is a basic solution.
It is easy to confuse yourself if you only choose to look at one component of the equilibrium and try to understand ...
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Another way to look at the problem:
If we add $\ce{H+}$ to the water through adding acid, then some of the $\ce{H+}$ would just remain in the solution as is, and some of them would react with $\ce{OH-}$ in the reaction $\ce{H+ + OH- -> H_2O}$. How much of the added $\ce{H+}$ reacts? To calculate this, you need to know that $\ce{[H+][OH-] = K_w}$ ...
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$n$-Factor is basically a method to find out a relationship between the compound and what it is equivalent to in terms of acidic nature or basic nature. Note that the two hydrogen atoms in $\ce{H2SO4}$ are both attached to oxygen. But in $\ce{H3PO3}$ two of the three hydrogens are attached to oxygen and the other hydrogen is directly attached to phosphorous.
...
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