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As the amounts of substance in the final solution are known to be $n(\ce{NH3}) = \pu{2 mmol},$ and $n(\ce{NH4^+}) = \pu{0.5 mmol},$ you may simply use the definition of the constant $K_\mathrm{b}:$ $$K_\mathrm{b} = \frac{n(\ce{NH4^+})[\ce{OH^-}]}{n(\ce{NH3})} = \frac{\pu{0.5 mmol}\times [\ce{OH-}]}{\pu{2 mmol}} = \pu{3.3E-5}$$ from where $[\ce{OH-}],$ $[\ce{... 4 All strong acids dissociate in water:$\ce{HX + H2O -> H3O+ + X-}$. Metals reacts with hydronium ions, not with a particular acid, e.g. like$\ce{M(s) + 2 H3O+ -> M^2+ + H2(g) + H2O}$. ( With nitric acid, it is more complicated due its reduction to$\ce{NO}$or$\ce{NO2}$, but it is not important now. ) Metallic cations and acid anions are not paired ... 3 Maybe your electrode is incorrectly calibrated. But it may be another effect : you should know that at high acidic concentration, the pH values is not obtained by taking the logarithm of the concentration of$\ce{H3O+}$. At high acidic concentration, the concentration should be replaced by the activity, which could be rather different from the concentration. ... 3 Lactic acid and lactate are part of an equilibrium reaction in aqueous solution. Even when the pH is far away from the pKa, there is still a small fraction of the minor species, and protonation/deprotonation reactions occur. If lactate were a bunny and the hydrogen ions in solutions were hats, lactic acid would just be the bunny after putting on the hat. If ... 2 This nomenclature comes from a long time ago. In the$19$th century, it was decided that the acidic compounds containing H, O and another atom X would be named from the name of this third atom X. If only one molecule of such an acid was known, the name would be made by adding the ending -ic. Example: Only one acid is known with a Carbon atom, and it is$\ce{...

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I have a solution which has $[\ce{H2SO4}]$ = 0.915M (I assume $[\ce{H+}]$ = 1,830 M). While $\ce{H2SO4}$ is a strong acid, $\ce{HSO4-}$ is not ($\mathrm{p}K_\mathrm{a}$ around 2). So your pH should be about zero, and you might not need to use the Hammett acidity at all.

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Your cited reaction proceeds, but in the reverse directions as follows: $$\ce{HCl (aq) + AgNO3 (aq) -> AgCl (s) + HNO3 (aq)}$$ So apparently, a weak acid can react to give a 'stronger acid', albeit, in dilute form, with the creation of an insoluble precipitate. This is relatedly evident, as one can also produce a dilute version of the acids like HI, HBr, ...

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Can a weak acid react to give a stronger acid? Well, yes, if you set up the right conditions. For instance, sulfuric acid (estimated first pK = -3) can be heated with $NaCl$ to boil off $HCl$ (pK = -5.9) according to Wikipedia. $HCl$ is driven from the reaction arena as a gas. You will claim that’s a special case, with unusual conditions (the boiling off). ...

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Neutralizations of any acid with oxides, hydroxides and carbonates always produce a salt and water. The only neutralizations that do not produce water are reactions of an acid $\ce{HX}$ with ammonia $\ce{NH3}$ or with organic derivates of ammonia, like amines. These reactions produce an ammonium salt $\ce{NH4X}$ and no water $$\ce{NH3 + HX -> NH4X}$$ If ...

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I think the below scheme depicts well how the intramolecular hydrogen bonding more favours the collapse of the carboxylic O—H bond in the orthohydroxybenzoic acid than in the parahydroxybenzoic acid.

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Wikipedia gives several methods to get boric acid. The first one uses a relatively common natural source of boron. Curiously compared with the other answer, boron trifluoride is excluded among the halides. Boric acid may be prepared by reacting borax (sodium tetraborate decahydrate) with a mineral acid, such as hydrochloric acid: $\ce{Na2B4O7·10H2O + 2 ... 1 Yes, hydrolysis of boron trifluoride. $$\ce{4 BF3 + 3 H2O → 3 HBF4 + B(OH)3}$$ The reaction commences with the formation of the aquo adduct,$\ce{H2O−BF3}$, which then loses$\ce{HF}$leading to formation of the products. The heavier trihalides do not undergo analogous reactions, possibly due to the low stability of the tetrahedral ions$\ce{BCl4-}$and$\ce{...

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