New answers tagged

0

Mixing equal volumes of 0.2M $\ce{HCl}$ and 0.6M $\ce{H2SO4}$ would yield a solution with nominal concentration of 0.1M $\ce{HCl}$ and 0.3M $\ce{H2SO4}$. Also given that $K_\mathrm{a2}$ for $\ce{H2SO4}$ is $1.2\times10^{-2}$. RANT- I have come to absolutlye hate problems that don't use significant figures consistently. Is the initial HCl concentration ...


2

Butyric acid also known under the systematic name butanoic acid is a carboxylic acid with the structural formula $\ce{CH3CH2CH2COOH}$ Source So butanoic acid and butyric acid are the same substance.


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The problem is that both titration curves are wrong: at the first equivalence points, the pH does not equal $\mathrm{p}K_\mathrm{a1}$ and the pOH does not equal $\mathrm{p}K_\mathrm{b1}$. As well, the two provided dissociation constants for carbonate ion, in the OP's upper figure, are also wrong. The figure below shows two titration curves: 1) for 10 mL of 0....


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The (equivalence)points, which you have marked in the graphs, are where all of the acid present in the sample has been neutralized by the base added with the titrant –and vice versa. But, pKa equals pH at the point where only half of the acid is neutralized. In other words: [HA] = [A-] -the concentration of acid equals the concentration of the corresponding ...


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I tried the Xanthan Gum method using muratic acid to clean the calcium leaching on my pool, which was bad, having sat for 7-8 months. I used 1 lb. of Xanthan Gum (about \$11 on Amazon) and about 3 gallons of water to make a thick gel. I then added about 1 gallon of it to a 2nd bucket and added muratic acid, 2 then a 3rd gallon to make it stronger. It just ...


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I upvoted the answer by @Poutnik, but here is a slightly different take. I assume the 98% and 37% concentrations are by mass, i.e., 100 g of 98% concentrated sulfuric acid contains 98 g of pure acid plus 2 g of water. I also assume the 98% and 37% are exact, to avoid initial fussing with significant figures. Round off to two digits is near the end. Take 100 ...


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The answer begins with the general calculation of mixing solutions with various densities. If you are interested only in the particular numeric results, or if you are scared by algebra, , skip the theoretical part between the horizontal lines. There is the mixing cross rule, deriving the ratio of mass of 2 mixed liquids from their mass percentage. $$\frac { ...


0

Not perfect, but here are a couple rules of thumb to start with for the most common cases. The key with both rules is to look at the molecular structure. A Bronsted acid typically has hydrogen attached to oxygen, sulfur or a halogen, all these being relatively electronegative elements. The hydrogen could also be attached to another nonmetal, especially ...


2

You are thinking too far ahead for a simple reaction. First of all, $\ce{I-}$ is not a very strong base to abstract a protone from alcohol. $\ce{I-}$ is the conjugate base of very strong acid, $\ce{HI}$. Yet, $\ce{I-}$ is a very good nucleophile. Therefore, the condition given ($\ce{NaI}$ in anhydrous acetone; $\ce{NaI}$ dissolves in anhydrous acetone and ...


2

The exothermic reaction of the acetic acid on $\ce{NaHCO3}$ provides heat and I do agree that the vigorous nature of the reaction forming $\ce{CO2}$ is likely instrumental in clearing the drain. Also, I suspect employing an excess of NaHCO3 is probably beneficial as on warming it releases CO2 and creating more alkaline Na2CO3, which will also attack grease. ...


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Quoting text from this source under the section "Why could baking soda and vinegar clean clogged drains?" : Bicarb soda and vinegar react because of the acid-base reaction. Bicarb soda is bicarbonate $(\ce{NaHCO3})$ and vinegar is acetic acid $(\ce{CH3COOH})$. When bicarb soda and vinegar react, they fizzle and sizzle and they expand. This is why ...


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$$\ce{NaHCO3(s) + CH3COOH(aq) -> CO2(g) + H2O(l) + CH3COONa (aq)}$$ From what I can see in the reaction, an amount of water and carbon dioxide is produced which has the potential to move around blockages in the drain similar to using club soda to remove stains from a shirt or table cloth. The resulting sodium acetate is a weak conjugate base and water ...


0

Ammonia is more basic than hydrazine if you look at the neighbors you will see $NH_3$, and $NH_2-NH_2$ where Ammonia has hydrogen as third neighbor where hydrazine have N as neighbors which gives more strong - I effect, after protonation.


2

When protonated, ammonia and hydrazine give their conjugated acids: $$\ce{NH3 + H3O+ <=> H4N+ + H2O} \tag1$$ $$\ce{H2N-NH2 + H3O+ <=> H3N^+-NH2 + H2O} \tag2$$ Let's rewrite these conjugate acids: $\ce{H3N^+-H}$ and $\ce{H3N^+-NH2}$ . We all know that electran withdrawing ability ($-I$ effect) of $\ce{-NH2}$ group is higher than that of $\ce{-H}$ ...


0

The sodium carbonate is just there to mop up any $\ce{HBr}$. The $\ce{Br2}$ adds across the double bond to give cinnamic acid dibromide. The treatment of cinnamic acid dibromide with $\ce{KOH}$ at elevated temperature eliminates 2 eq. of $\ce{HBr}$ to give Phenylpropiolic acid. This is known to decarboxylate through the intermediate acetylene anion.


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The Henderson-Hasselbalch Equation ignores the autoionization of water, but, it is affected indirectly since $\mathrm{p}K_\text{w}$ affects the value of $\mathrm{p}K_\text{a}$. The equilibrium between the acid disassociation constant and the ionic product for water is $\mathrm{p}K_\text{a} + \mathrm{p}K_\text{b} = \mathrm{p}K_\text{w}$.


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The equation $$\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = \mathrm{p}K_\mathrm{w}\tag{1}$$ has two degrees of freedom, so two values are independent on each other and the third one depends on the other two. Reaction equilibrium constant $K_\mathrm{a}$ for $$\ce{HA + H2O <=> H3O+ + A-}\tag{R1}$$ is chemically independent on $K_\mathrm{w}$ for ...


1

Chlorine is somewhat soluble in water: 14.6 g/100 mL H2O at 0 C, but drops to 0.57 g/100 mL at 30 C. So one way to reduce exposure to Cl2 would be to keep the temperature low and the volume adequate (dilution) to maintain all the chlorine in solution, with a little more volume for safety. There will still be chlorine escape, but this can be minimized by ...


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Fluorine ion and chlorine ion are very corrosive in nature, so In water they easily form the hydrohaloacid as order of acidic strength in water is $hi>hbr>hcl>hf$ hence the stability order will be $F^{-} >Cl^{-} >Br^{-} >I^{-}$ which also shows stability of their acids.


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$\ce{H2CO3}$ and $\ce{NaCl}$ do not react. The sodium chloride provides a surface with high friction which allows the $\ce{CO2}$ to transform into a gas. The same thing happens with mentos and fizzy drinks. One might say that the sodium chloride is serving as a catalyst for the transformation of $\ce{H2CO3}$ into $\ce{H+(aq)}$ and $\ce{CO2(g)}$.


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If we consider $\ce{HX}$ as the common formula for hydrogen halogenides, then there is an equilibrium $$\ce{HX(aq) + H2O <=> X-(aq) + H3O+(aq)}$$ that can be expressed also as: $$\ce{acid1 + base2 <=> base1 + acid2}$$ The equilibrium is competition of 2 acids, $\ce{HX}$ ansd $\ce{H3O+}$. $\ce{H3O+}$ is the strongest acid stable in aquaeous ...


3

Which biological and chemical processes underlie the effect of antiperspirants that make metal salts so effective? Aluminium-based complexes present in the antiperspirants react with the electrolytes in the sweat to form a gel plug in the duct of the sweat gland. The plugs prevent the gland from excreting liquid and are removed over time by the natural ...


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Chlorine bleach is commonly produced via the chloralkali process as a highly alkaline mixture of sodium hypochlorite (NaOCl) and hydroxide (NaOH) along with some unreacted NaCl as well. It's usually dangerous to lower the pH of chlorine bleach, as you may inadvertently generate chlorine gas $$ \ce{2H+(aq) + OCl- (aq) + Cl- (aq) \rightleftharpoons Cl2(g) + ...


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Your question: Can $\ce{NaOH}$ and $\ce{NaH2PO2}$ exist together in aqueous solution? I assume what you meant is your textbook is saying when two separate solutions of $\ce{NaOH}$ and $\ce{NaH2PO2}$ are combined, there should be no reaction. If that is what you meant "exist together," then your textbook is correct because there won't be an acid-...


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No, because osmosis is not an electrical phenomenon, but rather a chemical one. The concentration of protons would diffuse across the membrane if the fluid, I'm assuming which is water, will be allowed to interact with the other side of the membrane. However, water will not seek to increase the pH of an acidic solution within a membrane the same way it will ...


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Medical saline has a $\mathrm{pH}$ of 5.5 do to the dissolved $\ce{CO2}$ within the solution as well as factoring in the degradation of the PVC packaging. I reference the attached article from the International Journal of Medical Sciences from 2013: Benjamin AJ Reddi, “Why Is Saline So Acidic (and Does It Really Matter?),” Int. J. Med. Sci. 2013, 10(6), 747-...


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The inductive effect will always make the stronger acidic compound, especially with an sp hybridization as it will ultimately stabilize the electrophile best in solution. Though resonance will also contribute to the stability, a structure similar to allene will have more resonance as it will have more bonds focused in one area so close to the electrophilic ...


2

There is a very valid reason for using acid in batteries, where needed. You are right that salts are also good conductors but keep in mind they are not the best conductors. In aqueous solutions, proton is the best conductor known! Which ion is no. 2? Not surprisingly, the hydroxide ion. Look at the comparison, there is no match for proton in various tables.


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All depends on the used electrochemical system. The electrolyte is the part of this system, it is not just a passive medium to conduct the current. Some systems work in acidic environments like acid lead cells, some use salts like the Daniell cell, some use alkaline solutions like nickel-cadmium or nickel-metalhydride cells, and some do not even use water in ...


3

$\ce{BaSO_3}$ has the same property as $\ce{BaCO_3}$ : When reacting with an acidic solution, these two substances produce a gas which can get out of the solution, and displaces the equilibrium to the right : $$\ce{BaCO_3 + 2 H^+ -> Ba^{2+} + CO_2(g) + H_2O}$$ $$\ce{BaSO_3 + 2 H^+ -> Ba^{2+} + H_2O + SO_2(g)}$$ This allows these two substances to be ...


2

p-Nitrophenol does have a higher $\mathrm{p}K_\mathrm{a}$ than carbonic acid (7.15 vs 6.3), the two logarithms differ by less than one unit, if we render the acidic species in carbonic acid as $\ce{CO2}$ rather than $\ce{H2CO3}$. So the equilibrium constant in the reaction $$\ce{p-C6H4(NO2)OH + HCO3- <=> p-C6H4(NO2)O- + CO2 + H2O}$$ is close to $1$ ...


0

Usually there is some form of a redox reaction happening there, for example like $$\ce{M(s) + 2H+(aq) \to M^{2+} (aq) + H2(g)}$$ where $M$ is just some metal (like copper, iron, etc...) reducing hydrogen to its elemental/gaseous form.


0

I'd not concern about the heat increased by the dilution of $32\%$ $\ce{NaOH}$ with negligible amount of water. I'm telling this by my personal experience. In our organic lab program, our student workers have been diluting $50\%$ $\ce{NaOH}$ solution ($\pu{12.5 M}$) with water at room temperature to make $\pu{3 M}$ ($\ce{NaOH:H2O}$ is in $1:3.2$ ratio) and $\...


1

The solution gets warmer by diluting with extra water, but no violent reaction. I suggest to do a test, if the temperature raise is concerning you, e.g. because of the tubing temperature resistance. What can also be done is applying solution with graduating concentration, so warming would be spreaded in sevceral steps. But as hydroxide is already quite ...


1

Efficiency of a buffer is not an absolute parameter, but it is conditional and case dependent. If buffer differencial capacity ( $\frac{\mathrm{d}n}{\mathrm{dpH}}\cdot \frac{1}{V}$, where $n$ is molar amount of an added base/acid and $V$ is buffer volume) or integral capacity ($\frac{\Delta n}{\Delta \mathrm{pH}}\cdot \frac{1}{V}$ per $\Delta pH = \pm 1$) ...


0

Chloro group is considered a deactivating group according to Hammett plots of each possible reaction. The $\sigma_\mathrm{para}$ of $\ce{Cl}$-substituent is listed as $+0.227$ while $\sigma_\mathrm{meta}$ of $\ce{Cl}$ is listed as $+0.373$. It is known fact that $\sigma_\mathrm{meta}$ is an indicative of how much inductive effect contribute to the reaction, ...


0

Yes as you said the inductive effect of chlorine exceeds the strength of it's +R effect. This effect becomes clear when you study aromaticity, that chlorine is a deactivating group (i.e it destabilizes the carbocation formed) but is still Ortho - para directing, because it still causes stabilization by the +R effect in those positions. However unlike other +...


-1

Can't quite follow your equations but your friend is right. It takes twice as much $\ce{NaOH}$ as $\ce{H2SO4}$. In other words for every $\pu{1L}$ of $\pu{5M}$ $\ce{H2SO4}$, you need $\pu{2L}$ of $\pu{5M}$ $\ce{NaOH}$. The reactions are: $$\ce{2H+ + 2OH- -> 2H2O}$$ $$\ce{2Na+ + SO4^2- -> Na2SO4}$$ OVERALL: $$\ce{2NaOH + H2SO4 -> 2H2O + Na2SO4}$$


0

A proton does not react with another proton, as you state. It reacts with a metal. And this metal gives one electron to one proton. This makes a Hydrogen atom, then a Hydrogen molecule. Finally the metal becomes a positively charged cation, which passes into solution


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Tests for sodium hydroxide: Do a pH test. Sodium Hydroxide is basic. Do the flame test. A bright yellow flame shows that sodium ions are present. A few drops of dilute sodium hydroxide solution react to form a white precipitate with aluminium ions, calcium ions and magnesium ions. However, in excess sodium hydroxide solution The aluminium hydroxide ...


1

Assume concentration of solution A and B are $a\%$ and $b\%$, respectively (both are $\%(w/v)$). Suppose $a\% \approx b\% \approx 5\%$. Molar mass of ammonia ($\ce{NH3}$) is $\pu{17.03 gmol-1}$. Thus, if molarity of A and B are $M_A$ and $M_B$, respectively, then: $$M_A = \frac{a \ \pu{g} \ \ce{NH3}}{\pu{100 mL}} \times \frac{\pu{1 mol}}{\pu{17.03 g} \ \ce{...


1

Ortho tert butyl benzoic acid will be more acidic than the other one .because former is more bulkier so it will rotate the COOH group out of the benzene ring plane to a more extent . For NH2 group Ortho to CH3 or C(CH3)3 .Ortho effect never applies NH2 can never go out of the plane .rather this time CH3 and C(CH3)3 goes out of plane .so hyperconjugation ...


0

I really like the concept but the problem is to find a membrane that is permeable to water and H3O+ ions but impermeable in respect to the indicator. I understand you need something as in the picture below. Dialysis tubing are commonly used to prevent bigger molecules from crossing the membrane, but unfortunately the typical lower limit is for molecules of ...


1

For housewives, a solution is simply to add Epsom salt in equal doses to each mix and let the $\ce{Mg(OH)2}$ settle out of solution. Decant and repeat the experiment until one solution no longer produces any results of a white precipitate. The latter brand is weaker in its aqueous ammonia concentration based on the reaction: $$\ce{\overset{Epsom salt}{MgSO4 ·...


3

If we look at the classic case of ammonia and hydrogen chloride in the gas phase, the answer is yes -- but it's more complicated than one might think. The salt does form, but it requires either a proton-transfer medium such as water vapor or the formation of particles out of the gas phase to actually get the ions. The acid-base reaction is formally, $$\ce{...


0

The given solution is a buffer of the equilibrium: $$\ce{C6H5COOH + H2O <=> C6H5COO- + H3O+} \tag1$$ The buffer question can be easily solved using Henderson-Hasselbalch equation. In case you don't know how to derive it, we'll do it from the scratch. Following expression can be written from the equation $(1)$: $$K_\mathrm{a}=\ce{\frac{[C6H5COO-][H+]}{[...


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