Chemistry, 25.05.2021 23:00 AlmightyThadd

When an acetic acid solution is titrated with sodium hydroxide, the slope of the titration curve (pH versus volume of NaOH added) increases when sodium hydroxide is first added. This change shows that

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When an acetic acid solution is titrated with sodium hydroxide, the slope of the titration curve (pH versus volume of NaOH added) increases when sodium hydroxide is first added. This change shows that acetic acid is being converted to sodium acetate.

Explanation:

ABOUT TITRATION - Titration is a chemical analysis method for determining the amount of a constituent in a sample by adding an exact known quantity of another substance to the measured sample with which the desired constituent reacts in a specific, known proportion. A burette, which is essentially a long, graduated measuring tube with a stopcock and a delivery tube at the bottom end, is used to gradually administer a standard solution of titrating reagent, or titrant.

It is feasible to identify a good visual color indicator for many titration processes that will signify the end point at, or very near to, the equivalence point.

Acid-base titrations, precipitation titrations, complex-formation titrations, and oxidation-reduction (redox) titrations are examples of such titrations, which are categorised according to the nature of the chemical reaction that occurs between the sample and the titrant.

Titrations of metal ions with the reagent disodium are the most important titrations based on complex-formation processes (a salt of edetic acid, or EDTA).

Hence , the answer is that acetic acid is converted into sodium acetate.

Answer from: Quest

The net force is 100

Answer from: Quest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.

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When an acetic acid solution is titrated with sodium hydroxide, the slope of the titration curve (pH...

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