New answers tagged acid-base
2
Pure water dissociate according to the equation:
$$\ce{2H2O(l) <=> H3O+(aq) + OH- (aq)}$$
Assume that $K_\mathrm{w} = 1.00 \times 10^{-14}$ at room temperature. Thus, $\ce{[H3O+]}$ of solution is $\pu{1.00E-7 M }$. Now, if you add a trace amount of strong acid, this equilibrium would be disturbed and backward reaction occur to reduced some of added ...
1
Yes, it is definitely possible, but also relatively difficult as compared to electroplating zinc onto another metal.
It is a significantly important industrial process. Although aluminium itself is reactive, it develops an impervious oxide layer that quite much renders it unreactive.
However, comparatively larger negative standard electrode potential of ...
-1
Here is another way, using my old Excel spreadsheet. The equation comes from the relevant acid dissociation constants, mass balance, charge balance and water's auto-ionization, as per the reference in the spreadsheet screen shots. It is assumed that $\ce {pK_a} = 4.76$ for acetic acid. Be sure to look at the red text in the following screenshots.
The buffer ...
3
Suppose you have a $\pu{1.0 L}$ of acetate buffer solution made by $\pu{500.0 mL}$ of $\pu{1.0 M}$ $\ce{CH3COONa}$ solution adding to $\pu{500.0 mL}$ of $\pu{1.0 M}$ $\ce{CH3COOH}$ solution. Thus, $[\ce{CH3COOH}] = [\ce{CH3COONa}] = \pu{0.5 M}$. Therefore, according to Henderson–hasselbalch equation, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} = 4.75$ ($\mathrm{p}...
3
It can be prepared using hydrogen fluoride, in a double displacement reaction:
$\ce{SbCl5 + 5 HF -> SbF5 + 5 HCl}$
However, hydrogen holds onto fluorine more tightly than does antimony in $\ce{SbF5}$.
Consider the reverse of your proposed reaction, which is preferred energetically:
$\ce{SbF5 + H2 -> SbF3 + 2HF}$
0
In $\ce {SbF3}$, antimony has 3+ oxidation number. Oxidizing it with $\ce {F2}$ yields $\ce {SbF5}$ and antimony has 5+ oxidation number. Then adding $\ce{H2F2}$ just results in transfer of one fluoride ion, $\ce {F-}$, to $\ce {SbF5}$, resulting in $\ce {SbF6-}$ and $\ce {H2F+}$. There is no evident redox reaction if antimony trifluoride is simply mixed ...
3
Easy marks for phenolphthalein usability are:
$99.9\%$ of acetic acid is titrated at about
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{1000} = 7.75$$
$100\%$ of acetic acid is titrated at
$$\mathrm{pH} = 7 + \frac12 \cdot (\mathrm{p}K_\mathrm{a} + \log c) = 9.385 + \frac12 \log c$$
Phenolphthalein color transition is $8.2-10.0.$
3
Aluminum chlorohydrate (ACH) is a group of specific aluminum salts having the general formula $\ce{Al_nCl_{(3n-m)}(OH)_m}$ (Wikipedia). It is used in cosmetics as an antiperspirant and as a coagulant in water purification. Specifically in wastewater treatment processes, ACH is used as a coagulant to remove dissolved organic matter and colloidal particles ...
7
In theory, any molecule that contains a H atom ("proton") can act as a Bronsted acid - it can donate that proton. How likely this is to happen is described by that proton's pKa value, the lower the value, the more likely this is to happen (the more stable the conjugate base that forms will be). You can look up pKa values for different types of proton on an ...
0
I am first going to do the ‘flip side’ of the OP’s scenario and then basically flip it back. So we start with a 0.1 M aqueous solution of benzoic acid and the initial solution volume is 25 mL. Benzoic acid is a monoprotic weak acid with $\mathrm{p}K_\mathrm{a} = 4.20$. The initial pH is 2.606. The strong base used for the titration is 0.1 M NaOH. At the ...
0
The brown stain may be iron oxides that were dissolved from the stone and precipitated by the sodium bicarbonate, or possibly a sulfonation product of the asphalt which is not soluble in water, but coats the asphalt. Three techniques could reduce the staining:
1) Put a few spoonfulls of vinegar on the spot and let it stand several minutes before washing it ...
0
The first group in the periodic table has metals. Metallic oxides are predominantly basic. As we go across the period the basic property decreases, trend changing to first amphoteric then acidic elements.
The oxides mentioned above in the question do now fall into a particular group or period thus requiring brainstorming.
My suggestion is to convert the ...
9
You did the right thing by neutralizing with a base.
The best solution is to let time heal the stain. The surface is obviously damaged, but there is no way to reverse that damage. The stain will only go away if someone keeps on scrubbing the area until a new surface is exposed. The good news is that dirt, soil, and water will slowly remove and even scrub ...
3
Two rules that partially solve the problem:
1) Predominantly ionic oxides, such as those of magnesium and heavier alkaline earth metals, will be only basic in water, even if only weakly so due to limited solubility. You need significant covalent bonding of the metal to oxygen to get the metal to take on additional hydroxide ions in strong bases.
2) ...
0
Ringo has given an excellent answer for your questions and continued to direct you during comment session. I just want to tell you is, this question can also be answered by using Henderson–Hasselbalch equation.
Suppose after titrating, the equilibrium concentrations of BH is $\alpha$ and $\beta$ at pH 8.1 and 7.5, respectively. Thus, all [BH] comes from the ...
3
Sorry to tell you but your teacher's opinion is not right, just because the word "react" is not enough. Usually you label oxides as "amphoteric". You may say that if an oxide reacts with an acid and forms a salt or if an oxide reacts with a base forms a salt then you can say that it is an amphoteric oxide. Take the example of aluminum oxide, when dissolved ...
1
MOMCl reacts via an SN1 reaction and alcohol is the nucleophile, not an alkoxide. The amine serves to scavenge HCl. ADDENDUM: Typically, a generic alcohol in an inert solvent is treated with MOMCl 1 in the presence of a hindered tertiary amine such as Hunig's base 2. MOMCl, which is prone to ionization by virtue of the participation of the oxygen atom, forms ...
7
There are several interesting aspects to your question.
How are neutral amines effective bases in organic chemistry?
First of all, recall the definition of a base. Anything which accepts a proton (or donates a pair of electrons) is a base. Organic bases are very strong and corrosive! These bases should not be undermined, even though they may sound weak (...
1
Actually, if you balance given redox reaction strictly following the half-reactions method where you have to equate the number of transferred electrons, you won't end up with fractional coefficients:
$$
\begin{align}
\ce{\overset{+7}{Mn}O4- + 8 H+ + 5 e- &→ \overset{+2}{Mn}^2+ + 4 H2O} &|\cdot 2 \tag{red}\\
\ce{2 \overset{-1}{Cl}^- &→ \overset{0}...
2
The balanced equation seems correct. Multiplying by $2$ we get the following:
$$\ce{2 MnO4- + 16 H+ + 10 Cl- → 2 Mn^2+ + 5 Cl2 + 8 H2O}$$
which is also correct.
1
If you want a general solution for complex mixtures, where simplified formulas do not apply, you may need to involve numerical mathematics.
As solving the equation set may lead to polynomial equation of higher order you may like.
Enumerate equations of the known.
First, charge balance.
Then, mass balances.
Finally, dissociation equilibriums.
Involve a ...
0
[sorry for posting this as an answer, but it isn't possible to include an image when posted as a comment]
Do we also need to consider hydrogen bonding effects in the conjugate base, since the measurement is in water? Presumably forming the phenolate would increase the basicity of the amine through the inductive effect, and this could be stabilized by ...
2
I am afraid, direct acid-base titration is not the right way to analyze aspirin exactly because of the you stated - hydrolysis. The rule number no of any titration is that there should be no side reaction and it should go to completion almost instantly. The direct titration of aspirin is problematic because hydrolyzes pretty fast to salicylic acid- an ...
1
Answer to this question is, as orthocresol pointed out, simpler than what you think. As you know, $\ce{-Br}$ is deactivate the ring (even though it is o,p-directing) while $\ce{-NH2}$ is great ring activator. What that means is regardless of the mesomeric effect, compared to $\ce{-H}$, $\ce{-Br}$ is overall electron withdrawing and $\ce{-NH2}$ is electron ...
0
In this situation, considering the distance between the $\ce{-OH}$ group and the substituents we can ignore inductive effects and focus on the mesomeric ones. $\ce{-Br}$ has a $+M_s$ effect, same as the $\ce{-NH2}$ group. To make an $\ce{-OH}$ group more acidic we need electron withdrawing groups as the slight positive charge built on the oxygen atom will ...
1
The question
This is the question, corrected according to Poutnik's comment, and adding (i) and (ii) to distinguish the two scenarios:
$\ce{NH3}$ solution of $\pu{0.1 mol dm-3}$ is being added to a $\pu{25.0 cm3}$ of $\pu{0.1 mol dm-3}$ $\ce{HCl}$ solution. Calculate the pH of the solution when volume of added $\ce{NH3}$ solution is
(i) $\pu{25.0 ...
1
According to the Globally Harmonised System (GHS), hydrochloric acid with a concentration of c ≥ 25 % and sulfuric acid with a concentration of c ≥ 15 % are labelled "causes severe skin burns and eye damage" (H314) and are classified based on standard animal test data in skin corrosion Sub-category 1B ("Corrosive responses in at least one animal following ...
0
In the given molecule
At , D lone pair on oxygen is in conjugation with C=O group.Consequently it acquires a positive charge,its Kb is less for protoantion to take place.(as shown in scheme 1)
(as pointed by MIthoron)
At C lone pair on this ring (furan) are delocalised and is AROMATIC (as shown in Scheme 2). Comparatively is kb value is less for protonation ...
3
In short, none. As Mithoron said, the oxygen atoms can all act as Lewis bases and will all be protonated to an extent. If you rephrase the question to "Which of the atoms has the highest Kb?", well that's another story. Atom D has the lowest Kb, because the carbon in a carbonyl has a slight positive charge. Atom C would be difficult to protonate, the oxygen ...
4
Yes, concentrated strong acid can destroy litmus paper.
Paper is mostly cellulose, which is a carbohydrate polymer. Concentrated sulfuric acid will pull the hydrogen and oxygen out of it as water, leaving carbon behind. The indicator dye that's supposed to change color probably won't fare too well, either.
You can dilute the unknown by adding a drop or two ...
-3
Liquid pH testing solutions or drops are also available. Other indicators like phenolphthalein, methyl orange , methyl red can be used if you expect your test solution to fall in the pH range of the respective indicators. Also it will be wise to take a small sample of the solution to test it rather than testing the whole thing.
2
The rate of dissolution depends on $\mathrm{pH}$ and $\mathrm{pH}$ dependent equilibrium reaction chain :
$$\begin{align}
\ce{2 CH3COOH + CaCO3 v &<=> (CH3COO)2Ca + Ca(HCO3)2}\\
\ce{2 CH3COOH + Ca(HCO3)2 &<=> (CH3COO)2Ca + 2 H2CO3}\\
\ce{H2CO3 &<=> H2O + CO2(aq)}\\
\ce{CO2(aq) &<=> CO2(g)}\\
\end{align}$$
what ...
0
Depending on your stain, you might want to consider a soap as well. As for your more theoretical question, yes there is a threshold. If the concentration of acetic acid is high enough, it will be less effective. However, it would have to be very high. This is described by Le Chatelier's principle
1
I have found a paper which does this same experiment and provides a discussion of this phenomenon[1]. I will begin by just quoting from the discussion section:
It has been stated in the introductory section that a
maximum in the formal Cp° for a pair of thermally
stable free ions can be ascribed to an endothermic
change in the solvent shell around ...
1
When ammonium chloride is hydrolyzing, there is established equilibrium
$$\ce{NH4+ + H2O <<=> NH3 + H2O+}$$
with $$\mathrm{pH}=7-\frac 12 ( 4.75 + \log {c_{\ce{NH4Cl}}} )$$
$$\mathrm{pOH}=14 - \mathrm{pH}=7+\frac 12 ( 4.75 + \log {c_{\ce{NH4Cl}}} )$$
Concentration of ammonia is several orders higher than concentration of hydroxide ions, with ...
0
First, a brief refresher on $\alpha$ dissociation fractions. Citric acid, denoted $\ce{H_3Cit}$, is a triprotic weak acid. The successively less protonated citrate species are $\ce{H_2Cit^-}$, $\ce{HCit^{2-}}$, and $\ce{Cit^{3-}}$. Since all the citrate species are in the same volume of solution, molar concentrations (denoted by the usual square square ...
1
A vs. B
I will label your first reaction scheme, using the letters A-E for the deprotonated (i.e. as shown in the problem) form and the symbol $\ce{AH+}$ etc for the protonated form (after acting as a base and accepting a proton):
$$\ce{A (stronger base) + BH+ (stronger acid) <=> AH+ (weaker acid) + B (weaker base)}$$
Your argument about aromaticity ...
1
Initial note
The specific problem with your scenario is, that you need a buffer of buffer capacity progressively decreasing with increasing $\mathrm{pH}$, but capacity of simple buffers is for $\mathrm{pH} \lt \mathrm{p}K_\mathrm{a}$ increasing.
As consequence, its "breaking" capacity is too big, compared to "withstanding capacity".
General analysis
Let ...
0
Alkaline etching of glass tends to be less rapid than HF and leaves a smoother surface. This may not be what you do for the first etch on fresh glass, but it may smooth out the etch somewhat after an HF etch. The smoothing may also be caused by silicate ions that form in the alkaline etch fluid, so perhaps a solution of potassium (or sodium) silicate (ortho ...
1
For the 1st equivalence point, the tartaric acid $\ce{H2A}$ is titrated to $\ce{NaHA}$
For the solutions of acid salts, we can with advantage use the formula
$$\mathrm{pH}=\frac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{2}$$
that can be derived from equilibrium
$$\ce{2 HA- <=> H2A + A^2-}$$
For the 2nd equivalence point there is $\ce{Na2A}$...
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