# Tag Info

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.

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: $$\ce{CH3COOH + OH- <=> H2O + ... 6 I'm not sure I follow your logic. For a monobasic acid S (\ce{HA}) dissociation degree α is$$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 ...

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, ...

3

Strontium hydroxide is a strong base, so you can calculate $[\ce{OH-}]$ as $\pu{0.02 M}$, then use $$K_\mathrm{w} = \ce{[OH-][H3O+]}$$ $$1\cdot 10^{-14} = 0.02\cdot [\ce{H3O+}] \quad\to\quad [\ce{H3O+}] = \pu{5e-13 M} \quad\to\quad \mathrm{pH} = 12.3$$

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

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 ...

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 $\mathrm{p}H$ ...

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 ...

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 ...

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 $$K_\... 2 To be amphiprotic means that the chemical species can donate or accept H+ ions.$$\ce{H2CO3<=>[H+] HCO3^- <=>[OH-] CO3^{2-}}\ce{H3PO4 <=>[H+] H2PO4^- <=>[OH-] HPO4^{2-}}\ce{CH3COOH <=>[H+] CH3COO^- <=>[OH-] \text{No Change}}$$Thus since the acetate anion can't donate a proton, it is not amphiprotic. 2 Pure water (rain as well as distilled water) in equilibrium with the atmosphere (p_{\ce{O2}}=10^{-3.5}\ \mathrm{atm}) can be calculated to contain about$$\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\\ \...

2

I've fouled up this problem badly. Here is a more carefully thought out solution. So let's start at the beginning with the ICE table. Making the normal assumptions: Concentrations will be used instead of activities. The $\ce{Zn(CH3COO)_2}$ dissociates completely in aqueous solution, thus the $\ce{Zn^{2+}}$ and $\ce{CH3COO-}$ ions can be considered ...

1

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)} &+& \...

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, ...

1

The acidic environment would not allow creation of anions but anions of strong acids, that are too weak bases to be significantly protonated.

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{... 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 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 ... 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 }...

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 ...

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{... 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 ... 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+)$$The solution buffer capacity (not limited to presence of specific buffer ... 1 @ Bennett, I feel your pain. The first table you show is most unhelpful for students as they are trying to understand conjugate acid-base pairs. For what it's worth, I like to use a version of table 2 where I substitute, "Negligible acid/base" as the conjugate of a "Strong" (rather than "Not an acid/base"). I find this helpful as students understand the ... 1 If no specific activity coefficient data are available, I would use the general formula for ionic compounds of small enough ion cincentations:$$ \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 ...

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