New answers tagged acid-base
0
Taking into my consideration to the above comment, I solved the problem.
I used the following table to determine the amount of each species at the end of the reaction:
\begin{array}{c | c c c c c c}
\text{RXN} &\ce{Zn(CH3COOH)2}_\text{(aq)} & + &\ce{2NaOH}_\text{(aq)} &\ce{->}& \ce{2CH3COONa}_\text{(aq)} &+& \...
0
$\ce{SO4^2-}$ is actually going to make the pH basic because $\ce{HSO4-}$ is a weak acid (not $\ce{H2SO4}$, only the first hydrogen ionizes completely). By itself aqueous $\ce{Mg^2+}$ will not form magnesium hydroxide in water. But with the $\ce{OH-}$ from the $\ce{HSO4-}$ is will form a suspension with some dissolved and some not so dissolved, but still the ...
2
To be amphiprotic means that the chemical species can donate or accept H+ ions.
$$\ce{H2CO3<=>[H+] HCO3^- <=>[OH-] CO3^{2-}}$$
$$\ce{H3PO4 <=>[H+] H2PO4^- <=>[OH-] HPO4^{2-}}$$
$$\ce{CH3COOH <=>[H+] CH3COO^- <=>[OH-] \text{No Change}}$$
Thus since the acetate anion can't donate a proton, it is not amphiprotic.
1
Good question. If you show these three ions (as written) to a super intelligent being person who never took chem or physics, they will not be able to tell you which ion is amphoteric. In short, there is no simplr short-cut! This requires some knowledge of some formula writing conventions.
For simple inorganic ions, if H is written with a p-block element, ...
0
At the beginning of the titration, the solution is a weak base. At the equivalence point, the solution is a weak acid (with some more spectator ions). The pH of these solutions changes with dilution, both in the direction of neutral pH.
In contrast, at the half-equivalence point (or midpoint) of the titration, the solution is a 1:1 buffer. As long as the ...
3
Strontium hydroxide is a strong base, so you can calculate $[\ce{OH-}]$ as $\pu{0.02 M}$, then use
$$K_\mathrm{w} = \ce{[OH-][H3O+]}$$
$$1\cdot 10^{-14} = 0.02\cdot [\ce{H3O+}] \quad\to\quad [\ce{H3O+}] = \pu{5e-13 M} \quad\to\quad \mathrm{pH} = 12.3$$
1
A relatively strong acid that does not employ borane, cyano, or halogen functions, containing only carbon, hydrogen and oxygen, is croconic acid. The structure of this acid from Wikipedia is given below:
Croconic acid has two stages of dissociation, given in Wikipedia as $pK_{a1}=0.80, pK_{a2}=2.24$. The second stage dissociation especially stands out, ...
0
$$\ce{Lets assume a general salt BHA (a salt of weak base and weak acid)}$$
$$\ce{(Here B^+= NH_{4}^+ and HA^-= HCO_{3}^-)}$$
$\ce{BHA}$ dissociates in water as follows:
$$\ce{BHA => B+ + HA-}$$
The following three main reactions take place:-
$$\ce{B+ + H2O <=> BOH + H+}$$
$$\ce{HA- + H2O <=> H2A + OH- (HA- acting as base)}$$
$$\ce{HA- <=&...
2
At the point of equivalence, when the added molar amount of $\ce{HCl}$ is just matching the present molar amount of $\ce{CH3NH2}$, there is solution of the weak acid $\ce{CH3NH3+}(+\ce{Cl-})$, that I will denote as $\ce{HA}$.
If we neglect $\ce{H+}$ from water dissociation and consider $c_\ce{A-}\ll c_\ce{HA}$, then $c_\ce{A-} = c_\ce{H+}$
Therefore
$$K_\...
0
pKa value comparsion:
2-florophenol > 2-chlorophenol > 2-bromophenol > 2-Iodophenol
Lower pKa is stronger acid but for this substances pKa ranges 8 ~ 9 results:
1. These aren't strong acid
2. 2-Iodophenol is stronger
3. Don't ask this question again simply find substance's pKa
0
The reason behind your professor's statement is that when you have an acidic solution, the H+ is always around, and therefore it quickly reacts with any negative charges that it meets. That's why you can't have any negative charges, they won't last long!
Remember that proton exchange is among the fastest chemical reactions too. I hope this helps!
1
The acidic environment would not allow creation of anions but anions of strong acids, that are too weak bases to be significantly protonated.
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The questions asks you to differentiate Na2SO4 and Na2SO3. There are two main concepts involved here:
H2SO3 is a very weak acid (better to write SO2 (aq)) and H2SO4 is an extremely strong acid. Certainly, HCl would not be able to displace H2SO4. Another point which you should remember is that salts of strong acids and strong bases are pH neutral. However, ...
1
The last statement
…sodium sulfate and sodium sulfite are both basic and would turn pH indicator the same color
is true only for $\ce{Na2SO4}$ as it's formed by both strong base and strong acid and won't noticeably affect pH.
$\ce{Na2SO3}$, on the other hand, undergoes hydrolysis:
$$
\begin{align}
\ce{Na2SO3 + H2O &<=> NaOH + NaHSO3} \\
\ce{...
2
Pure water (rain as well as distilled water) in equilibrium with the atmosphere ($p_{\ce{O2}}=10^{-3.5}\ \mathrm{atm}$) can be calculated to contain about
$$\begin{align}
\mathrm{pH}=-\log[\ce{H+}]&=5.65\\
-\log[\ce{HCO3-}]&=5.65\\
-\log[\ce{CO3^2-}]&=10.3\\
-\log[\ce{H2CO3^*}]&=5.0\\
-\log[\ce{CO2}]&=5.0\\
-\log[\ce{H2CO3}]&=7.8\\
\...
0
The order of acidic nature would be:
Water > ethyne > ammonia> ethane.
Since oxygen has highest electronegativity amongst the atoms of the compounds so it's hydrogen will be most acidic and hence will be best acid among them . Ethyne contains sp carbon and as a result the electronegativity of the carbon in ethyne is more than $sp^3$ due to increased s- ...
0
The aryl group is electron withdrawing in nature, so as the no.of of aryl groups increase, the acidity increase while $\ce{CH3}$ being electron donating in nature leads to +I effect and thus to basic strength..
That's why (IV) will come first then (I) then (II) and then after (III).
So the order as per decreasing basic strength is (IV)> (I)>(II)> (III)
-1
The blue color was caused by copper being stripped from the wire. btw, when your hair turns orange, it is because it is being stripped of color, among other other things (chemical bonds), which will cause it to be brittle and also absorb any hair dye to the extreme that it feels waxy...I'd suggest you cut off a piece of hair for a test before dunking your ...
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These systems are more complex than they might seem at first. I more or less agree with Martin's answer above but there are some subtleties at play. There are two acids present which can react with added hydroxide ions: the CH3COOH molecules and the hydronium ions.
Initially, as OH- is added, it reacts with and removes the very small amount of hydronium ions ...
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As I mentioned in the comments above, chemicalize.com is a cheminformatics website which is the kind of resource you are looking for.
You can use the search tool to find your compounds. The input is the structure itself, which you can draw in-situ.
Here is the $\mathrm{p}K_\mathrm{a}$ data I retrieved from the aforementioned site. Note that these are ...
1
Why is an atom with high polarizability a strong nucleophile?
An atom with a higher polarizability will share his electrons more easier since they are less attracted by the nucleus. The more the electrons are near the nucleus, the more it's hard the share them with an electrophile. $\mathrm{I^-}$ is thus more nucleophile than $\mathrm{F^-}$.
As you'll se ...
18
Equilibrium will be far to the right, as a stable six membered ring is formed. The proton speaks for itself.
References: Queen, A. The kinetics of the reaction of boric acid with salicylic acid. Can. J. Chem. 1977, 55 (16), 3035–3039 DOI: 10.1139/v77-421.
3
By googling "boric acid salicylic acid"
I have found the salicylic acid acts like it was a diol toward the boric acid, but just 1:1.
https://www.nrcresearchpress.com/doi/abs/10.1139/v77-421
1
First of all, within the Lewis framework, you may have substances that can act as both acid and base (though most of the time as one or the other, not both at the same time).$^{1}$ The adjective for such substances is "amphoteric".
It comes down to a bit of chemical intuition to decide which is which. It is true that almost anything that has a lone pair can ...
2
There is no special unit analysis.
It's a recommended practice to solve the problem algebraically first using proper notations for physical quantities, and plug the numeral values at the end minding the units — this way you reduce the chance of making the erroneous calculations and keep track of all units; as a bonus, you simplify handling significant ...
1
The Unit/dimension analysis is in this case rather "using a cannon against sparrows".
The first spotted error is $\pu{500 g}$ of $\ce{HCl}$. Where did it come from ?
Note that the molar mass of $\ce{HCl}$ is about $\pu{36.5 g / mol}$.
Get the molar mass of $\ce{NaOH}$.
Calculate the amount of mols of $\ce{NaOH}$.
Calculate the equivalent amount of mols of ...
1
Edit: At the equivalence point, the solution contains the dissolved salt $\ce{CH3NH3Cl}$, dissociated to $\ce{CH3NH3+ + Cl-}$, as @MaxW noted.
You can calculate $pH$ of the conjugate acid $\ce{CH3NH3+}$ at its dissociation equilibrium.
$$\begin{align}
K_w &= K_a . K_b = 10^{-14}\\
pK_w &= pK_a + pK_b = 14\\
\ce{ CH3NH3+ &<=> H+ + CH3NH2 }...
3
Although pH scale is rather strictly used for aqueous solutions, the concept of pH in organic or non-aqueous solutions is not trivial. It is called apparent pH or operational pH when organic solvents are present. There is a practical problem here as well, do you think your buffers will dissolve in ethanol? Once you figure this out, you can read a section in ...
0
Here, you need to learn about following factors:-
1. Hydrogen Bonding with solvent and intermolecular and
2. Ortho/Para effect due to resonance
3. Steric Inhibition Effect
The first thing which you have to do is make the conjugate bases and check their stability.
Solution to Q1-The anion is stabilised due to the resonance effect due to which the electrons ...
1
@ Bennett, I feel your pain. The first table you show is most unhelpful for students as they are trying to understand conjugate acid-base pairs. For what it's worth, I like to use a version of table 2 where I substitute, "Negligible acid/base" as the conjugate of a "Strong" (rather than "Not an acid/base"). I find this helpful as students understand the ...
1
If no specific activity coefficient data are available, I would use the general formula for ionic compounds of small enough ion cincentations:
$$
\begin{align}
I &= \frac{1}{2}\sum_{{\rm i}=1}^{n} c_{\rm i}z_{\rm i}^{2} \\
\log{\gamma} &= -0.509 . ( z_+^m . |z_-|^n )^{\frac{1}{m+n}}. \sqrt{I} \quad (*)\\
\end{align}
$$
$c_i$ is the ion ...
0
I think Andrew's (see above answer) initial choice for the source of carbonate as sodium bicarbonate ($\ce{NaHCO3}$) is correct based on its enthalpy of solution, which is listed as $\pu{+18.7 kJ\:mol^{-1}}$ in literature (Ref.1). I disagree with his final choice of sodium carbonate ($\ce{Na2CO3}$) based on its solubility in water as the carbonate of choice ...
0
Much more practical to flush the water heater occasionally as recommended by the manufacturers ( in the US). Keep in mine a normal water heater also has a magnesium /zinc/aluminum anode that will dissolve in acids. However citric, formic and acetic acids are used to cleans steels with minimal damage. PS : Refineries will pay you a fortune for your HCl ...
2
Several organic acids are endothermic when dissolved in water: citric and tartaric are two that I have experience with. They would be good candidates for experiment 4.
The difficult one is the carbonate. I think ammonium carbonate is endothermic (the chloride and nitrate are). This could be the solid in experiment 2. It would have an odor of ammonia, which ...
2
The obvious choice for the source of carbonate is sodium bicarbonate (UPDATE: sodium bicarbonate is not soluble enough per other answer, but sodium carbonate is a possibility), which has a positive enthalpy of solution even at high concentrations, so let's call that B.
The acid is a little trickier. If you look up the standard enthalpy of solution of most ...
0
There are two equilibria to consider, $\ce{ 2H2O <=> H3O^+ + OH^-}$ and $\ce{ HA +H2O <=> H3O^+ + A^-}$ where HA must be a weak acid, and for example, NaA its sodium salt. Because of the equilibria there is no real distinction between solutions of (1) pure HA, (2) mixtures of HA and NaA and (3) pure NaA.
In your question one solid must be a ...
6
$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$
My question is, which acid (CH3COOH or the conjugate H3O+) will the base KOH react with?
You have to satisfy two equilibria, the one you wrote and the auto-dissociation of water.
$$\ce{H3O+ + OH- <=> 2 H2O}$$
If you add up the two reaction, you get a third one:
$$\ce{CH3COOH + OH- <=> H2O + ...
1
Zhe's comment is not wrong, but I'd say that your teacher isn't either.
Consider a 0.1 molar solution of acetic acid. The water will be about 55 molar.
The twist here is that the proton of the hydronium ion is very liable in water. In other words, the proton exchange between water molecules happens very very fast. That is why using a mineral acid makes a ...
0
I was going to mark this as too broad, but it's not too broad, because the answer is quite simple: in general, you cannot determine the acid or base properties of a substance from its formula.
As a concrete example, consider the formula $\ce{C4H7NO3}$.
Here are two possible isomers for that formula:
The left compound is an acid. The right compound is a ...
6
I'm not sure I follow your logic.
For a monobasic acid S $(\ce{HA})$ dissociation degree $α$ is
$$α = \frac{[\ce{H+}]}{c_\mathrm{a}},$$
where $c_\mathrm{a}$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_\mathrm{b}$:
$$c_\mathrm{a}V_\mathrm{a} = c_\mathrm{b}V_\mathrm{b} \implies ...
1
Weak bases, that are not salts, do not dissociate, but partial react with water as
$$ \ce{R-NH2 + H2O <<=> R-NH3^+ + OH-}$$
$\ce{KOAc}$ is not weak base, but a salt.
The weak base is $\ce{OAc-}$, that is created by dissociation of the salt $\ce{KOAc}$ in water, which undergoes protonization:
$$\begin{align}
\ce{KOAc &-> K+ + OAc-} \\
\ce{...
0
Any aqueous solution increases its buffer capacity toward $\mathrm{p}H = 0$ or $\mathrm{p}H = 14$, as more and more acid or basis is needed to change $\mathrm{p}H$ by a given interval. It happens even without presence of any buffering substance, aside of strong acids or bases.
If an acid, monoprotic or diprotic, is added, a peak(s) of maximum buffer ...
1
Can I multiply Ka1 and Ka1 to eliminate [$\ce{C4H5O6−}$], and then get the concentration of C4H4O62− necessary by plugging in 0.1 M for [C4H6O6] and the target pH in the appropriate form in [H+]
No, because $\ce{C4H5O6−}$ is a one of the major species. In fact, if you add the tartaric acid and its double salt at equimolar ratios, $\ce{C4H5O6−}$ will be the ...
0
I would take different approach.
From given $pH$ and acidity constants, calculate ratios of the particular forms of the tartaric acid.
From given total molar amount, you get the molar amounts for the particular forms.
From molar amount of particular forms, you can get molar amount of a hydroxide to be added to the tartaric acid, or of a strong acid to be ...
0
Part A
- Calculating the initial $\pu{pH}$(before adding $\pu{0.02 mol}$ of $\ce{HCl})$
The equilibrium here is
$$\ce{CH3CH2NH2 + H2O <=> CH3CH2NH3+ + OH−}$$
For $\ce{ CH3CH2NH2} , \mathrm{K_b} = 4.6\times{10^{−4} }$
$$\mathrm{K_b} = 4.6\times{10^{−4} }=\frac{[\ce{CH3CH2NH3+}][\ce{OH−}]}{ [\ce{CH3CH2NH2}]}$$
The nominal concentrations of the ...
1
Generally, the maximum buffer capacity is at $\ce{pK_a}$ . The tartaric acid is somewhat special for 2 reasons:
It is a diprotic acid with both $\ce{pK_a}$ very close, with the $\ce{pK_{a1}}$ rather low, being affected by the reason 2. :
$$\ce{pK_{a1}}=2.89,\ce{pK_{a2}}= 4.40 (L+)$$
The solution buffer capacity (not limited to presence of specific buffer ...
2
If we consider $\ce{HA}$ as a weak acid, then at the half equivalence point, $$\mathrm{p}H = \mathrm{p}K_\mathrm{a}$$ As $$\mathrm{p}H = \mathrm{p}K_\mathrm{a} + \mathrm{p[\ce{HA}]} -\mathrm{p[\ce{A-}]}$$ and for the half equivalence point,
$$\mathrm{p[\ce{HA}]} =\mathrm{p[\ce{A-}]}$$
So the higher the $\mathrm{p}K_\mathrm{a}$ is, the higher is $\mathrm{p}H$ ...
5
At first glance, use of oxalic acid dihydrate ($\ce{H2C2O4.2H2O}$ or simply OADH) as a primary standard seems really odd to anybody including me (although I'm not a analytical chemist). I'd also expect a primary standard to be oven dried and cool it in desiccator before use as we all did in our college analytical lab using potassium hydrogen phthalate (KHP). ...
5
Jander-Blasius (14. Ed., 1995) uses nonhygroscopic sodium oxalate, dried at 230-250°C (it decomposes above 250 according to wikipedia), to standardise permanganate titer solution against. They give no other recommended standard for manganometry, so I assume this is it.
I have no idea why anybody would want to use (or recommend using) the free acid instead, ...
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