New answers tagged

1

As mentioned by @orthocresol, just comparing methane and acetylene shows a difference of $\approx 20$ $\mathrm pK_\mathrm a$ units and the amount of resonance required to overcome this difference is very high. Comparing $\mathrm p K_\mathrm a$ of phenol($10.0$) and methanol($15.5$), we see that the presence of one resonating ring only produces a 5.5 $\mathrm ...


1

There are few conceptual and factical points: There is no such molecule in water solutions like $\ce{NH4OH}$. Nitrogen atom is not able to form 5 covalent bonds. There is either hydrated ammonia $\ce{NH3(aq)}$, either hydrated ammonium and hydroxide ions, following equilibrium: $$\ce{NH3(aq) + H2O <<=> NH4+(aq) + OH-(aq)}$$ respectively $$\ce{NH3(aq)...


12

You are most likely getting an inaccurate value. As anticipated in Poutnik's comment above is difficult to reliably measure a $\mathrm{pH}$ of very acidic solution but the $\mathrm{pH}$ scale is indeed open, and negative $\mathrm{pH}$ are a real thing. It is easy and more meaningful to state a very high concentration or activity rather than reporting ...


0

So is it 0.01 mol, not 0.01 M ? In such a case molarity of the acetate does not matter, what does matter is its molar amount. But I would rather say it IS a typo and they meant 0.01 M as 0.01 mol/L. For that case, The acid volume and acetate molar amount matter. By other words, for either of cases, a particular acetate solution molarity is not required. What ...


2

Question: Determine the $\mathrm{pH}$ of the solution resulting when $\pu{100 cm^3}$ of $\pu{0.50 mol dm-3}$ $\ce{CH2ClCOOH}$ is mixed with $\pu{200 cm^3}$ of $\pu{0.10 mol dm-3}$ $\ce{NaOH}$. I'm not sure what level of chemistry is OP's in, but the given solution for the question is for the chemistry students with appreciable knowledge of stoichiometry of ...


1

$\ce{CH2ClCOOH}$ is a weak acid; $\ce{NaOH}$ is a strong base. Now, tell me, what do you think happens when you mix a strong base with a weak acid? They react. $$\ce{CH2ClCOOH + NaOH -> CH2ClCOONa + H2O} $$ A neutralisation reaction in aqueous medium depends upon the number of equivalents of acid and base. $$N_1V_1=N_2V_2 $$ At this stage, an equivalence ...


2

This was answered in the comments, but these second equilibrium expression should not actually be a $K_b$ value. Rather, it should be $1/K_a = 55600$. Using this value in the same calculations, we find that $x=5.5805\times 10^{-4}$ and $y = 5.5795\times 10^{-4}$, so that $\ce{pH} = -\log(x-y) = 7.00$.


2

Actually, there are three reasons why BF3 is more acidic than AlCl3. BF3 is smaller in size and can easily attract the incoming pair of electrons. Fluorine is more electronegative than chlorine. So boron has less ionisation potential as compare to aluminium thus forms anion easily. Boron is more electronegative than aluminium, therefore boron has more ...


-2

Lewis acids are those which can accept electrons easily. Now, in BF3 molecule, B is small in size and thus can easily attract the incoming pair of electrons. Also, the presence of three fluorine atoms makes the boron acquire a partial positive charge, due to which boron easily accept electrons acting as Lewis acid. But, in the case of AlCl3, the size of Al ...


6

I think it would produce something like "LiOH2," You are missing the charge. It should be something like $\ce{LiOH2+}$, which is an unconventional way of writing $\ce{Li+(aq) + H2O(l)}$. So where is my misunderstanding? You have to get used to interpreting the (aq) designation after ionic compounds. To help you out, here is the equation written ...


1

No, adding acid to a ferrous sulfate solution will not prevent its being oxidized by bromine water. Fresh FeSO$_4$ solutions are light blue-green, almost colorless, but pick up O$_2$ readily: Fe$^3$$^+$ + e$^-$ --> Fe$^2$$^+$ +0.770 V and O$_2$ + 2 H$_2$O + 4e$^-$ --> 4 OH$^-$ +0.401 V (The half cell voltages are only indicative of what's ...


2

The $\ce{Na+}$ ion is a spectator ion in this case. It doesn't participate in the chemical reaction. It is found unchanged on both the reagent and product sides (red top-right and green bottom-right reaction formulas), so it can be canceled out to get the net reaction (red left). From there, he did the usual acid-base calculation to get a formula for $\ce{Kb}...


0

Sodium ion is the weak conjugate acid of strong base i.e. NaOH. So the show which one ot of the two ions furnished undergoes hydrolysis to a greater extent the Na+(which does not react) is crossed and hydrolysis of stronger base C2H3O2- to produce OH- ions is shown.


1

PubChem Magnesium ascorbate $(\ce{Mg(C6H7O6)2})$ shows a monobasic salt. Wikipedia says pKa's are 4.10 (first), 11.6 (second) In addition to Poutnik's answer, the crystal ionic radii in pm according to wikipedia of $$\ce{Mg^2+ is 86pm and Ni^2+ is 83pm}$$ This following coordination complex formation reaction is known $$\ce{Ni^2+(aq) + 2(dmgH2)->Ni(dmgH)2(...


2

$\mathrm{p}K_\mathrm{a}$ relation in the answer of Kav Wikipedia says pKa's are 4.10 (first), 11.6 (second) gives the hint a monobasic salt with alkali metals should be very mildly alkalic, as pKa2 is farther from pH 7. $pH \approx \frac{\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2}}{2} = \frac{4.10+11.6}{2}=7.85$ OTOH, it will probably form a complex ...


0

Acids are hydrogen ion donors. When the react with water, they can give a hydrogen ion to form $ \ce{H3O+}$. For example: $ \ce{HCl(aq) + H2O(l) \rightarrow Cl-(aq) + H3O+(aq)}$ Simple acids, such as $ \ce{HCl}$ or $\ce{H2SO4}$, can be recognized as acids by the H at the start of the formula. Other more complex acids may be written with $ \ce{COOH}$ at ...


1

H3O+ is just the combination of a a $\ce{H+}$ ion, which we know to be released from the dissociation of an acid in an aqueous solution, with a water molecule. It happens because the $\ce{H+}$ is so, so positive (and therefore so reactive) that the water molecules, with its lone pairs (which are locally negatively charged), are willing to form a dative ...


3

What is meant by a "very strong" Be−O bond? If berylium's tendency to hold on to water ligands is unusually strong, is it due to its small ionic size? You got that absolutely right. $\ce{Be\bond{-}O}$ is a strong bond because of the small size of Be. Smaller cation size means a stronger pull on the $\ce{O}$ electrons, thus reducing the bond length ...


0

Since your weak acid is neutralized to the end point by a strong base, $\ce{NaOH}$, the final reding does not depend on the percentage of dissociation of the weak acid before neutralization. Suppose your ionization of the weak acid is as follows: $$\ce{HA + H2O <=> H3O+ + A-} \tag1$$ Thus at the end of neutralization: $$\text{amount of $\ce{NaOH}$ ...


2

Short answer is: Yes and no. It depends on what acid-base theory is applied. Redox reactions like $\ce{Mg -> Mg^2+ + 2 e-}$ are not usually considered acid-base reactions, but over-generalized acid-base theories like Usanovich acid-base theory treat them as a special case of acid-base reactions. Within this theory, $\ce{Mg}$ is relatively strong ...


1

As @Maurice pointed out, the result of a neutralization reaction is not impacted by the dissociation percentage. Is $\pu{[}\ce{HA}{] = 10^{−3} M}$ or is $\pu{[}\ce{H+}{] = 10^{−3}}$? $\pu{[}\ce{HA}{] = 10^{-3}}$. Always remember, neutralization in aqueous medium depends upon the number of equivalents of acid and base: $$N_1V_1=N_2V_2 $$ Since $V_1=V_2$, ...


1

First, you should be given $K_\mathrm{a}$ of the weak acid. If you know it, you can set up the expression for $K_\mathrm{a}$: $$\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}= K_\mathrm{a}$$ Let $x$ be the amount of the weak acid dissociate at equilibrium. Thus, at equilibrium, $[\ce{H+}] = [\ce{A-}] = x$ and $[\ce{HA}]$ will be its initial concentration minus $x$. ...


2

Inexpensive [about US\$0.30 / liter] marble or limestone scraps, $\ce{CaCO3}$, likely would serve well. You could make an outer bucket with polyethylene netting, so that the pickling bucket would be well below the level of surrounding limestone chips. Coarse netting would not work with powders. Make the neck just large enough to admit the parts. Better would ...


3

There is more to it than meets the electronegative eye. Get into the habit of transferring attributes into energy. In this case the relevant energies are the electron affinity of the base money ($\ce{RO}$ vs $\ce{RS}$) versus the bond energy of the moiety to hydrogen. A strong anionic Bronsted-Lowry base will have a low value for the first versus a high ...


0

Assuming the water is at the same temperature and pressure on both sides, the only way to create a difference in pH would be through the presence of conjugate acids and/or bases other than $\ce{H_2O}$, $\ce{H^+}$ and $\ce{OH^-}$. I.e., you can't have pure water at the same temperature and pressure on both sides, yet have a difference in pH. [Pressure would ...


2

No, that's not how it all works on more than one level. First, the amount of $\ce{NH4+}$ out there is small. And by "small", I mean really, really small - about as small as you can imagine, only smaller. Many times smaller, I must reiterate. If some substance was to react with it (which is certainly not outside the realm of possible) and form a ...


3

Dry ferrous sulfate heptahydrate is green(ish) and is expected to make a green(ish) aqueous solution. Anhydrous ferrous chloride is described as green to yellow, and the dehydrate and tetra hydrate are green to blue-green (CRC Handbook). Now color is broad and continuous (see the visible spectrum), not digital, like how many fingers am I holding up. This ...


2

I have asked OP to verify the solution concentration but didn't get the answer. Thus, I assume it is $8\% \ (w/w)$. Thus, if you assume $[\ce{HA}] = c$ then: $$ c = 8\% \ (w/w) = \frac{\pu{8 g}\text{ of HA}}{\pu{100 g}\text{ of sol}} \times \frac{\pu{1.0 mol}\text{ of HA}}{\pu{60.05 g}\text{ of HA}} \times \frac{\pu{1.01 g}\text{ of sol}}{\pu{1.0 mL}\text{ ...


2

I think your teacher's version of hydrolysis of sodium peroxide is straight out from Wikipedia article, which follows the reference 1. Wikipedia states that: Sodium peroxide hydrolyzes to give sodium hydroxide and hydrogen peroxide according to the reaction: $$\ce{Na2O2 + 2H2O -> 2NaOH + H2O2}$$ However, even though it can seemingly be considered to go ...


5

$\ce{HO2-}$ is a weaker base than $\ce{NaOH}$. but $\ce{O2^2-}$ is much stronger base than $\ce{HO2-}$ and does not occur in water solutions in significant amount. But its salts ( sometimes in the form of hydrates ) can be precipitated at highly alkaline solutions of hydrogen peroxide. Additionally, lack of product presence supports the respective ...


Top 50 recent answers are included