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Please forget about equivalents and normal solutions ! It is an old theory, that was abandoned in the middle of the $20$th century. It has been replaced by moles, molarity, and similar parameters. A solution containing $n$ moles of a solute in one liter water has a concentration expressed in mole per liter, that can be printed on the label of the flask. The ...


3

You could simply calculate the $\mathrm{pH}$ of the solution. $K_\mathrm{a{_1}}$ of $\ce{H2CO3}$ is $\pu{4.30E-7}$ and $K_\mathrm{a{_2}}$ of $\ce{H2CO3}$ is $\pu{4.72E-11}$.The approximate $\mathrm{pH}$ of the solution can be given by: $$\mathrm{pH}=\frac{(\mathrm{p}K_\mathrm{a{_1}} + \mathrm{p}K_\mathrm{a{_2}})}{2}$$ This yields $\mathrm{pH}= 8.347$, which ...


2

An aqueous solution of potassium bicarbonate is acidic ... towards a suitable base. The existence of a dissociable proton in the anion guarantees that hydroxide ion could act as such a base, accepting the proton from the bicarbonate ion; but says nothing about neutral water. It turns out that neutral water is not so strong a base as to pull the proton off ...


2

Spectator ions in any kind of reaction are those that do not take part in the interaction. In precipitation reactions, they are those not forming the least soluble salt. In acid-base reactions, they are those not manifesting in the given environment either acidic, either basic behaviours. Typically cations of alkali metal, or alkali earth metals, or anions ...


1

Aeration does not seem to be the way to go. If there are no other ions in the water, electrolysis (the Kolbe reaction: https://en.wikipedia.org/wiki/Kolbe_electrolysis) would convert the $HOAc$ to ethane and carbon dioxide: $2CH_3CO_2^- --> 2e^- + 2 CO_2 + H_3C-CH_3$ This process could probably be arranged to be done in a pipe with electrodes on the sides,...


1

Acetic acid does not form an azeotrop with water and is less volatile than water. So aeration of water would do the opposite - enriching of water by acetic acid. Reverse osmosis should help, perhaps after neutralization to be mostly in acetate form. The question is, if it is worthy the troubles. The cheaper, easier and faster way could be replacing the ...


1

This procedure from an MP-AES manufacturer features copper analysis (among other elements) without the use if hydrochloric acid. Hydrogen peroxide is used in this procedure but would not affect the extraction of copper. A copper standard is also available in "5% nitric acid" solution from the same company. So the people who are selling this ...


1

Just to add some more information for you with some clarification. I gather that you are asked to calculate the equivalent weight of nitric acid here. $$\ce{Cu + HNO3(aq, dil) -> Cu(NO3)2 + NO(g) + H2O(l)}$$ So first thing to keep in mind that the equivalent weight is determined with respect to a single role of the molecule of interest. An exception is a ...


1

What happens in salt hydrolysis is enhancement of water ionization with salts. Try asking yourself where do $\ce{H+}$ and $\ce{OH-}$ come from in the solution of $\ce{AB}$, and you'll find that they come from $\ce{H2O}$. Or equivalently, hydrolysis of $\ce{A-}$ can be written as $$ \ce{A- + H+ <=> HA,H2O <=> H+ + OH-}\stackrel{\text{add}}{\...


1

When the pH is equal to the pKa, you have high concentrations of both conjugate acid and conjugate base. You need the weak base to react with added acid, and the weak acid to react with added base to stabilize the pH. When the pH is very different from the pKa, one of those two species will be at very very low concentration (there is an exponential ...


1

Why does a buffer work best at the $\mathrm{pH}$ closest to its $\mathrm{p}K_\mathrm{a}$? For the novices like you who just begin to learn chemistry, this question can be easily explain by using Henderson Hasselbalch equation. Suppose you have an aqueous weak acid solution ($\ce{HA}$). It will be in ionic equilibrium as follows: $$\ce{HA (aq) <=> H+ (...


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