# Tag Info

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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 ...

7

$$\ce{NaHCO3(s) + CH3COOH(aq) -> CO2(g) + H2O(l) + CH3COONa (aq)}$$ 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 ...

7

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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I think we should not conflate "solubility" with "dissociation". They are different phenomena. The former has to do with a molecule breaking away from the other solute molecules and forming interactions with the solvent. The latter is a process which occurs after dissolution, whereby the solute molecules break apart to form ions, ...

5

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 ...

4

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. $$\frac { ... 4 If c is the concentration of \ce{Mg^2+} in the \ce{Mg(OH)2} solution, the concentration [\ce{OH-}] = 2c. At \mathrm{pH}~9, [\ce{OH-}] = \pu{1E-5 M}, then c = \pu{5E-6 M}. So you have to dissolve \pu{1.25 μmol} \ce{Mg(OH)2} in \pu{250 mL} water. This is \pu{71.3 μg} of \ce{Mg(OH)2}. This is difficult to do in practice, as such a small ... 4 There is an easy way to do what OP wants, assuming OP wants to prepare solutions at \pu{25 ^\circ C}. So, OP can prepare saturated solution of \ce{Mg(OH)2} solution:$$\ce{Mg(OH)2_{(s)} <=> Mg^2+_{(aq)} + 2OH-_{(aq)}}$$Since K_\mathrm{sp} of \ce{Mg(OH)2} is \pu{5.61 \times 10^{-12} M3}, you can find the solubility of \ce{Mg(OH)2} at \pu{... 3 Whether magnesium hydroxide is a strong or weak base is a matter of disagreement. Usually it is considered a weak base because of its low solubility. Whether it's a strong or weak base in terms of dissociation is not completely known. IUPAC reports a single pK_b value, generally between 2 and 3, based on hydrolysis measurements; but because the ... 3 I upvoted the answer by @Poutnik, but here is a slightly different take. I assume the 98% and 37% concentrations are by mass, i.e., 100 g of 98% concentrated sulfuric acid contains 98 g of pure acid plus 2 g of water. I also assume the 98% and 37% are exact, to avoid initial fussing with significant figures. Round off to two digits is near the end. Take 100 ... 3 According to the Brønsted-Lowry acid base theory, an acid is a compound that releases a \ce{H+} ion to give a conjugate base. For most acids, this reaction exists in a dynamic equilibrium. The equilibrium constant of this reaction is what defines K_\mathrm{a}. Generally more stable the product, the more of the forward reaction takes place. Using this ... 3 \ce{Mg(OH)2} is a strong base since it is ionic in nature; it usually dissociates completely and so it's degree of dissociation is one. For weaker salts, the concentration values you assigned for the ions, x and 2x respectively, do depend on the molarity of the \ce{Mg(OH)2}. You have to define x in the terms of its degree of dissociation (\alpha),... 3 The equation$$\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}$$is chemically independent on K_\mathrm{w} for ... 3 Chlorine bleach is commonly produced via the chloralkali process as a highly alkaline mixture of sodium hypochlorite (NaOCl) and hydroxide (NaOH) along with some unreacted NaCl as well. It's usually dangerous to lower the pH of chlorine bleach, as you may inadvertently generate chlorine gas$$ \ce{2H+(aq) + OCl- (aq) + Cl- (aq) \rightleftharpoons Cl2(g) + ...

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 ...

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 $1$ ...

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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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. ...

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 $\ce{-H}$ ...

2

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 ...

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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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$K_\mathrm{a}$ is the measurement of equilibrium constant when an acid dissociates as: $$\ce{HA <=> H+ + A-}$$ $$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$ The term $\mathrm{p}K_\mathrm{a}$ is defined as $\mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a})$. So lower the $\mathrm{p}K_\mathrm{a}$, the better is the acidity. In the example given, if ...

1

Now much base of any kind you need to get $\mathrm{pH} = 7$? None, the water itself is enough. For $\mathrm{pH}$ levels near 7 (8 is pretty much eligible) you want to account for the $\ce{OH-}$ ions that come from water. $[\ce{OH-}] = 1 \times 10^{-6} (\mathrm{pH} = 8)$ $[\ce{OH-}][\ce{H+}] = 1 \times 10^{-14}$ (water at room temp) $[\ce{OH-}] = [\ce{H+}] + ... 1 The problem is that both titration curves are wrong: at the first equivalence points, the pH does not equal$\mathrm{p}K_\mathrm{a1}$and the pOH does not equal$\mathrm{p}K_\mathrm{b1}$. As well, the two provided dissociation constants for carbonate ion, in the OP's upper figure, are also wrong. The figure below shows two titration curves: 1) for 10 mL of 0.... 1 Chlorine is somewhat soluble in water: 14.6 g/100 mL H2O at 0 C, but drops to 0.57 g/100 mL at 30 C. So one way to reduce exposure to Cl2 would be to keep the temperature low and the volume adequate (dilution) to maintain all the chlorine in solution, with a little more volume for safety. There will still be chlorine escape, but this can be minimized by ... 1 Medical saline has a$\mathrm{pH}$of 5.5 do to the dissolved$\ce{CO2}$within the solution as well as factoring in the degradation of the PVC packaging. I reference the attached article from the International Journal of Medical Sciences from 2013: Benjamin AJ Reddi, “Why Is Saline So Acidic (and Does It Really Matter?),” Int. J. Med. Sci. 2013, 10(6), 747-... 1 The solution gets warmer by diluting with extra water, but no violent reaction. I suggest to do a test, if the temperature raise is concerning you, e.g. because of the tubing temperature resistance. What can also be done is applying solution with graduating concentration, so warming would be spreaded in sevceral steps. But as hydroxide is already quite ... 1 Efficiency of a buffer is not an absolute parameter, but it is conditional and case dependent. If buffer differencial capacity ($\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\$) ...

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