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5

$\ce{CF3OH}$ is more acidic because of the inductive effect of the $\ce{-CF3}$ group: Fluorine is very electronegative and therefore it pulls the electron-density toward itself weakening the $\ce{O-H}$ bond in $\ce{CF3OH}$. Chlorine is less electronegative than fluorine, thus $\ce{CCl3OH}$ is less acidic than $\ce{CF3OH}$. Because $\ce{H}$ is even less ...


-3

PH = - log 10 [H3O+] [H3O+] = -antilog 10 (PH) [H3O+] = - 10^6.99 Because antilog b (x) = b^x Therefore , [H3O+] = 9772372.21


1

One problem is there is not enough voltage for this thing to sell. Say you have molar acid and 1 molar base, both strong and both functioning as ideal solutions. Thermodynamically you have one electron transfer associated with the net reaction $\ce{H^+}_{aq}+\ce{OH^-}_{aq}\to\ce{H2O}$ and the equilibrium constant is $10^{14}$ at room temperature. Put ...


1

$\mathrm{pH}$ is essentially a convention. It is defined as $$-\log_{10} [\ce{H+}]$$ since the concentrations of the solutions commonly used lie in the interval $$[10^{-14}\ \mathrm{mol/L},1\ \mathrm{mol/L}]$$ and thus the $\mathrm{pH}$ lies in $$[0,14]$$ But nothing constrains an aqueous solution from having a $\mathrm{pH}$ that does not lie in this ...


2

Since $p_\ce{CO2}$ is $\pu{30 mmHg}$, and we are given the $0.03$ of $\frac{\pu{mM}~\ce{CO2}}{\pu{mmHg}~\ce{CO2}}$ ratio, we can calculate the concentration of the $\ce{CO2}$ in plasma. $$\pu{30mmHg}~\ce{CO2}\times 0.03\frac{\pu{mM}~\ce{CO2}}{\pu{mmHg}~\ce{CO2}}= \pu{0.9 mM}~\ce{CO2}$$ $$\ce{CO2 + H2O \xrightarrow{\text{carbonic anhydrase}} H2CO3}$$ $$\...


2

For a phosphate buffer with $\mathrm{pH} = 7,$ the two dominant species are $\ce{H2PO4-}$ and $\ce{HPO4^2-}$. The relevant $\mathrm{p}K_\mathrm{a}$ is $7.2$ (this is $\mathrm{p}K_\mathrm{a2}$). From the Henderson-Hasselbalch equation, using $\mathrm{pH} = 7,$ the ratio of $\ce{HPO4^2-}$ to $\ce{H2PO4-}$ is about $0.631.$ You can see this in the alpha diagram ...


1

This particular question is testing basic knowledge of buffers. In the most simplest explanation possible, a buffer will resist changes in pH the best when the pH is at the pKa. In the case of phosphoric acid, there are three acidic protons thus making three pKa values. Pulling the following table from Wikipedia $\hspace{3.25cm}$ We can see $pK_{a2}$ is ...


8

The key approximation made in deriving the Henderson-Hasselbalch equation is that the equilibrium constant can be written as $$K=\frac{c_{H^+}c_{A^-}}{c_{HA}}$$ that is, we assume activity coefficients are unity. If you take the base-10 logarithm of this equation and rearrange terms you arrive at the HH equation:$$pH=pK_a+\log_{10}\left(\frac{c_{A^-}}{c_{HA}}...


3

For the derivation of the Hendersson-Hasselbalch equation, it is assumed that both hydronium and hydroxide ions are minor species, i.e. there concentration is low compared to that of the buffer species. In your exercise, the hydronium ion concentration (according to your calculation) is about 0.03 M or higher. With a initial concentration of $\ce{H2SO3}$ of ...


2

A balanced equation must have a mass balance i.e., the masses should be equal on both sides and charges must be balanced as well. Oxidation number is just a way of book-keeping. Nothing fundamental there. $$\ce{Fe^2+ + Cr2O7 + 14H+ -> 2 Cr^3+ + Fe^3+ 7 H2O}$$ What are you forgetting? It is mass balanced. Does $\ce{Cr2O7}$ exist? Hint: It is potassium ...


2

The colour of neutral red in aqueous solution changes from violently red (protonated cation) to yellow/mostly colourless (the uncharged amine). If the $pKa$ is 6.8, then the dye is still very much red at pH 6.8, and you won't be able to see the colour difference to, say, pH 5. You have to keep adding base up to a pH of eight until the concentration of the ...


-4

Here resonance is destabilizing as resonance here would mean that two oxygens will have negative charge and that would destabilize the conjugate base.


1

The given purine has two structures as shown below. In $\ce{9H–Purine}$ (figure 2), lone pairs are far apart from each other (3 and 7). Comparatively in $\ce{7H–Purine}$ (figure 1) lone pairs are very near each other (3 and 9) . Nearness of lone pairs in $\ce{7H–Purine}$ could leed to greater repulsions between them. This could make it energetically ...


2

Deprotonation creates a negative charge on the phenolic oxygen. The better you can stabilise this charge, the more favored this process is going to be. You are right in saying that the charge can be stabilised by delocalisation in the aromatic ring. However, the electrons of the methoxy oxygen can also be delocalised in the ring (to a much lesser extend, of ...


3

I believe have found a answer. It's valid mathematically, make sense physically but I don't know if chemically is true. I posted to community appreciation. There we go! The reactions ionization of weak acid: $$\ce{HA + H2O <=> H3O+ + A-}\qquad K_\ce{a}=\frac{\ce{[H3O+][A-]}}{\ce{[HA]}} \tag{1} \label{eq:KWeakAcid}$$ ionization of weak base: $$\...


22

On negative oxidation states, in general Although it's usually a topic that's covered relatively late in a chemistry education, negative oxidation states for transition metals[1] are actually quite alright. On the Wikipedia list of oxidation states, there are quite a number of negative oxidation states. Some textbooks have tables which only show positive ...


0

Well, you can explain it as a lewis base. In ammonia, nitrogen forms 3 single covalent bonds with 3 atoms of hydrogen. However it has 5 electrons in its valence shell. Which means that there are two electrons left a.k.a lone pair of electrons. Hence, it acts as a lone pair electron doner making it a lewis base.


6

In compound $\ce{A}$ , lone pair on $\ce{N}$ is in conjugation with two $\ce pi$ bonds (as shown below). Therefore this lone pair is delocalized (resonance structures in the figure below) to a greater extent. In compound $\ce{B}$ , lone pair on $\ce{N}$ is in conjugation with a single $\ce pi$ bond , while the other pi bond is not in conjugation or is ...


5

Vinegar Test: No Visible Change After Two Hours As advised above, I decided to test vinegar on an inconspicuous spot. First I created a pattern of exposed paint using masking tape. The shape was just something random that I thought I could most easily recognize, even if just a slight difference in the car's finish: Then I placed vinegar-soaked cloth on ...


40

The $\mathrm{pH}$ of pure water (rain as well as distilled water) in equilibrium with the atmosphere ($p_{\ce{CO2}}= 10^{-3.5}\ \mathrm{atm}$) can be calculated as follows. $$[\ce{H2CO3^*}]=K_\mathrm H\cdot p_{\ce{CO2}}$$ where $[\ce{H2CO3^*}]$ is the total analytical concentration of dissolved $\ce{CO2}$, i.e. $[\ce{H2CO3^*}]=[\ce{CO2(aq)}]+[\ce{H2CO3}]$, ...


9

Under atmospheric pressure, dissolved carbon dioxide can reach an equilibrium state in water that yields a pH of as low as 5.7


16

You are forgetting an important component of the air: carbon dioxide. When it dissolves in pure water (=rain water), it makes it acidic. It is not considered that harmful. Acid rain has a negative connotation; it is mainly caused by anthropogenic activities. The low pH of acid rain is due to sulfur oxides and nitrogen oxides and it is indeed below 5.7. ...


6

Your problems are caused by using the Arrhenius theory, which is based on electrolytic dissociation. According to the Arrhenius definition, acids are compounds that dissociate and release hydrogen ions $(\ce{H+})$ into the solution: $$\ce{HCl -> H+ + Cl-}$$ Bases are defined as compounds that dissociate and release hydroxide ions $(\ce{OH-})$ into the ...


2

To illustrate Oscar's answer further, it should be noted that boron, despite existing in more than ten allotropic modifications, is in general quite chemically inert (especially in crystalline form). For example, boron doesn't react with hydrogen directly and all boranes are synthesized by other means. Although boron may be oxidized by fluorine at room ...


4

We might as well consult the experts, namely storage tank manufacturers. This site describes several types of storage tanks, none of them metal except for rubber-lined steel, in which (we hope) the rubber keeps the acid out of contact with the steel. Rubber lining of steel is often used in processes such as acid rinsing tanks where the steel is needed for ...


2

Who said boron is a base? The fact that it does not weaken or neutralize most acids ought to tip you off that it isn't. Boron is slowly oxidized when powdered and exposed to nitric acid (see here), and we may call it "basic" by virtue of the products being much weaker acids than the nitric acid. But usually and with better precision the boron (or any ...


1

Think of oxidation in a classical sense for a minute or so. Oxidation is addition of oxygen to a molecule or it can also mean removal of hydrogen. If you carefully look at the structure of ascorbic acid, you will see that the two -OH groups are now converted into carbonyl groups. Those two hydrogens can be lost as a proton, H+.


0

Draw the ions of respective acids. See which one becomes more stable. Nitric acid $\ce{N}$ is in a bad state with a positive charge. it's a highly eletronagative atom near oxygen becoming negative will become more stable. See also what happens with $\ce{H3PO4}$. $\ce{P}$ is not eletronegative as $\ce{N}$ so it doesn't care much as $\ce{N}$ for $\ce{O}$ ...


1

How about heating a metal hydroxide such as magnesium hydroxide? At sufficiently high temperature the hydroxide decomposes according to the reaction $\ce{Mg(OH)2 -> MgO + H2O}$ which is most simply interpreted as a proton transfer from one hydroxide ion to another: $\ce{2 OH^- -> O^{2-} + H2O}$ This "disproportionation" would not take place at ...


1

The $\ce{O^{2-}}$ dianion is very unstable, especially in water. There is too much electronic density for the nuclear charge of oxygen to hold. We generally do not consider $\ce{OH-}$ to be amphoteric because, except in very extreme conditions, it will not give its proton away.


2

How a titration curve is affected when a poorly soluble salt is formed? If by "I think that titration curve don't be affected" you mean that the graph of pH vs volume of strong base will not be affected by the precipitations, then you are right. This is because the species that form the precipitate are not part of any other equilibrium reaction. how to ...


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