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9

Because the acid and conjugate bases are equimolar - and because the equilibrium constant is small, obviating the need to solve a quadratic if you approached this in another way - the Henderson-Hasselbalch equation is: $$\pu{pH} = {\rm p}K_\mathrm{a} + {\rm log}_{10}\left({[A^-]\over[HA]}\right)$$ $$\pu{pH} = 4.74 + {\rm log}_{10}\left({0.2\ {\rm M}\over 0.2\...


6

The strength of permanganic acid that you quote, combined with that of potassium hydroxide as a base, would guarantee that pure potassium permanganate is neutral in aqueous solution. But commercially prepared potassium permanganate is made in the presence of alkali, the use of potassium instead of sodium arising from the fact that the reaction scheme does ...


4

$ \begin{align} (n_{\ce{NaOH}})_i= \pu{0.2 mmol}\\ (n_{\ce{HCl}})_i= \pu{0.1 mmol}\\ (n_{\ce{NaH2PO4}})_i= \pu{0.1 mmol}\\ (n_{\ce{Na3PO4}})_i= \pu{0.05 mmol} \end{align} $ At first the $\ce{NaOH}$ will react with $\ce{HCl}$ as per the following reaction: $$ \begin{align} \begin{array}{ccccc} \ce{NaOH} & + & \ce{HCl} & \ce{->} & \ce{NaCl} ...


2

Ionic product of $\ce{H2O}$ is changing with temperature. It is $14.94$ at $0$°C, $14.17$ at $20$°C, $13.83$ at $30$°C and $12.26$ at $100$°C. So the pH of pure water is $7.47$ at $0$°C, $7.08$ at $20$°C, $6.92$ at $30$°C, and $6.13$ at $100$°C


2

The reason for this is that the hydrogen atom attached to the carbon will not be lost. Only the hydrogen attached to the oxygen will be ionized. In fact, in a solution of formic acid, not all molecules will be ionized anyways. Only very strong acids will fully ionize. Even in the case of sulfuric acid, not all molecules will fully ionize and lose the second ...


1

[OP] This gives a net 0 (carboxy) + 1 (amino) + 1 (side chain) = +2 charge. This is approximately correct. See DavePhD's answer for a more accurate treatment. [OP] Why do sites such as this say that at $\mathrm{pH} = 2$, lysine's charge is only +1, not +2? The table you cite is for proteins. When lysine gets incorporated into a protein, it forms peptide ...


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