New answers tagged

1

Nylon 612, primarily used for filaments and toothbrushes, is resistant to clove oil at 20 °C [1, p. 2034]: Nylon 12, used for coatings, is also resistant to clove oil at 20 °C [1, p. 1829]: The tech data on nylon at newmantools.com also suggests there is no effect of clove oil on nylon in general. References Chemical Resistance of Thermoplastics; ...


2

Neutral oxides do not manifest either acidic either alkalic behaviour ( but eventually very weak in extremely alkalic/acidic environment). Amphoteric oxides manifest, more or less balanced, both behaviours in reasonably strong acidic/alkalic environments.


0

"I am trying to understand why Nitrogen in Ammonia is a proton acceptor." I am going to modify your question slightly: it is a proton acceptor in aqueous solution. Water, being the solvent of life and being very common on Earth, is generally the most pertinent solvent. Ammonia is a proton acceptor, or if you prefer an electron-pair donor, in water. "...


5

Carbonic acid does not exist either in powder either in liquid form near normal conditions.. ( It can be created and detected at special kryogenic or gaseous phase conditions not applicable for its usage as "on shelf acid".) In a pure form, it exists only as its salts: bicarbonates ( of alkali metals ) and carbonates. As an acid, it exists at ambient ...


2

A simple way to think about this is as follows (assuming we are talking about an aqueous solution). The dissolution of $\ce{CO2}$ in water is followed by three chemical reactions: $$ \begin{align} \ce{CO2 + H2O &<=> H2CO3}\label{rxn:R1}\tag{R1}\\ \ce{H2CO3 &<=> HCO3- + H+}\label{rxn:R2}\tag{R2}\\ \ce{HCO3- &<=> CO3^2- + H+}\...


0

First you consider the state of hybridisation .all are SP3 .resonance is only possible with P orbitals and d orbitals too .here lone pairs of nitrogen in all are in SP3 back bonding is possible between 2p-2p ; 2p-3p ; 2p -3d . So as this cases are no longer valid we have to consider the electronegativity as you told in question .but remember it is my point ...


4

Let's suppose we want to titrate a solution containing an unknown monoprotic and weak acid. We use a strong base, such as $\ce{NaOH}$. When the number (and moles) of hydroxide ions is equal to the amount of hydronium ions, here we have the equivalence point. The equivalence point is, when the molar amount of the spent hydroxide is equal the molar amount ...


3

The reaction you wrote down is wrong on two counts. The reactant is not a hypothetical tetraaqua complex and the product is not a hypothetical tetraammin complex. The correct reaction is as shown below: $$\ce{[Cu(H2O)6]^2+ (aq) + 4 NH3 (aq) <=> [Cu(NH3)4(H2O)2]^2+ (aq) + 4 H2O (l)}\tag{1}$$ Note that I have used an equilibrium arrow here: the ...


0

From the comments: [OP] Okay Thank you all I finally got it. so K is just thrown out the window. and we are left with the equation (F- + H2O <-> FH + OH-) So i take the Ka to make Kb by (Kw/Ka) and my concentration of F- and OH- was (0.02/0.3) which I plugged into sqrt((Kb)*(M of OH- or F-)) took the -log of that got the pOH and then converted it back ...


10

A protonated carbonic acid (trihydroxymethyl cation) $\ce{H3CO3+}$ does exist, and is in fact incredibly stable in the solutions of superacids [1] up to $\pu{0 °C}.$ One way of obtaining $\ce{H3CO3+}$ is dissolving inorganic carbonates and hydrogen carbonates in magic acid at $\pu{-80 °C}$ [2]: $$\ce{CO3^2- or HCO3- ->[FSO3H-SbF5/SO2][\pu{-80 °C}] ...


-1

$\ce{H3CO3}$ does not exist. Why should it exist ? Where does this formula come from ? Of course you are allowed to write any formula you want. It is not forbidden to write $\ce{HC4O3}$ $\ce{H3CO5}$ $\ce{H3C2O3}$ or any other combination of C, H and O atoms that you may think of. Just for fun ! But Nature ignores your fantasy. Nobody has ever been able to ...


0

So, in principle, it's possible to make a buffer without adding the salt? Yes and no.... The math is pretty simple. Let's say that there is a monoprotic acid $\ce{HA}$. $$\ce{HA <=> H^+ + A-}$$ $${\mathrm{K}_\mathrm{a}} = \dfrac{\ce{[H+][A-]}}{\ce{[HA]}}$$ let $c$ represent the overall concentration of HA species. If $$\ce{[A-] = [HA] }$$ then $$\...


0

Interestingly, H2O2 has been considered an acid because its dissociation constant is smaller than that of H2O: "The available literature indicate that hydrogen peroxide is a weak inorganic acid with a dissociation constant between 2.24 and 2.4E-12 at 25 °C, which corresponds to a value for pKa ranging from 11.62 to 11.65. Hydrogen peroxide - Registration ...


0

Metal hydroxides are basic in water. Non-metal oxides are acids in water. The limit between metals and non-metals is a sort of staircase going through the periodic table from the middle of the first line to the lower corner at the right-hand side. Aluminium touches this staircase. This is why Aluminium behaves like a metal and like a non-metal, and why ...


4

In this answer a hydrocarbon anion called Kuhn's anion has the formula $\ce{C_{67}H_{39}^-}$ has reported $pK_b=8.1$ which would correspond to the neutral hydrocarbon having $pK_a=5.9$. The anion, together with several hydrocarbon cations with which it forms stable salts, is shown below (taken from the answer referenced above; primary reference J. Org. Chem....


1

Phenols have higher acidities than aliphatic alcohols due to the aromatic ring's tight coupling with the oxygen and a relatively loose bond between the oxygen and hydrogen of the hydroxyl group directly attached to the aromatic carbocyclic nucleus. For this reason, electron withdrawing groups (EWG) and/or electron donating groups (EDG) attached to aromatic ...


0

I understand the procedure but not the questions. So there are two reactions: $$\ce{2HCl(aq) + CaCO3(s) -> H2CO3(aq) + 2 Cl-(aq) + Ca2+(aq)}$$ $$\ce{HCl(aq) + NaOH(aq) -> H2O(l) + Cl-(aq) + Na+(aq)}$$ The first reaction has reactants in non-stoichiometric amounts. The second reaction, a titration, is done with stoichiometric amounts, yielding a ...


2

The thing you have to realise is that H-bonding can lower the ease with which the acidic $\ce{H}$ is removed, but it does not render it completely 'locked' within the molecule. What you have to understand is, the hydrogen bond shown here is not that strong as compared to the other kinds of H-bonds which have been observed in chemistry Let us consider some ...


1

$\mathrm{p}K_\mathrm{a}$ is inversely proportional to acidity of a compound similar to $\mathrm{p}K_\mathrm{b}$ which is inversely proportional to basicity as it is $-\log K_\mathrm{a}$ and $-\log K_\mathrm{b}.$ In this case nitro group is a electron-withdrawing and deactivates the phenyl group the most, whereas $\ce{Cl}$ is a moderate deactivating group ...


1

The pH is a quantity that is measured in an aqueous solution by dipping a glass electrode in the solution. It is equal to -log[H+] in dilute solutions. For concentrations > 0.01 M, the pH deviates from log [H+], because the concentration [H+] should be calculated by dividing the number of moles H+ by the volume of the free water (and not the volume of the ...


4

Acidity or acid strength can be expressed either in terms of the proton affinity (PA) of the conjugate base, e.g., PA($\ce{A-}$), or in terms of the acid dissociation constant, e.g., $\mathrm{p}K_{\mathrm{a}(\ce{HA})}$. The former expression is often favored among gas-phase chemists, whereas the latter expression is most familiar to solution chemists. In ...


3

To begin with, removing another proton from anionic species is not very favorable. For example, the monohydrogen phosphate dianion is not very acidic $(\mathrm{p}K_\mathrm{a}\sim 12)$ despite it having an $\ce{OH}$ group. In the case of $\ce{H3PO2},$ the second and third protons are bonded directly to $\ce{P},$ which is much less electronegative than $\ce{O}...


2

You are right: 1 cubic meter of water contains $\pu{1E-4}$ moles of $\ce{H+}$ ions. But one mole is $\pu{6E23}$ atoms or ions. So, $\pu{1E-4 mol}$ contains $\pu{1E-4}\times\pu{6E23} = \pu{6E19 ions}.$ Here we are.


-1

One possible explanation for this is hyperconjugation due to the attached alkyl chain, as shown below: since the acetate ion has one more $\alpha$-H atom than the propionate ion, the acetate ion is more stable than the propionate ion in the gaseous state and has more tendency to donate the hydrogen atom, making it more acidic.


2

So your solution is already at a pH equal to the pKa of the weak acid you are about to add. Your hunch is that the pH remains the same after adding acid to a solution. [...] we will end up with what we wanted: The same concentration of the weak acid and it's conjugated base. It depends on the concentration of what is already in solution and of what you ...


4

Nitrogen is more electronegative than hydrogen hence N will pull electrons towards itself. Hence not a good candidate to donate electrons! Yet it makes bonds with hydrogen, as do other electronegative elements (think of hydrogen chloride and water). So polar bonds form in all kind of molecules. lets take the case of Oxygen O=O, here each Oxygen atom has ...


1

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'...


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 ...


4

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 ...


8

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 ...


Top 50 recent answers are included