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## New answers tagged ph

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

8

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\... 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 ... 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

The autoionization equilibrium of water is satisfied irrespectively of $\mathrm p\ce{H}$. That's why $\mathrm p\ce{H}$ and $\mathrm p\ce{OH}$ add up to a constant: $$K_\mathrm{w}=a_{\ce{H3O+}}a_{\ce{OH-}}$$ $$\rightarrow \mathrm p K_\mathrm{w}=\mathrm p\ce{H} + \mathrm p\ce{OH}$$ This is a bit different from the typical scenario in which Le Chaterlier's ...

4

Recall that $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$ Where $\ce{[OH-]}$ is the concentration of hydroxide ions. Suppose a solution has $10^{-7}$ moles of $\ce{[OH-]}$ ions. Then this implies $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$ and $\mathrm p\ce{OH} = -\log_{10}\space [10^{-7}]$ which is equal to $7$. Suppose a solution has $10^{-4}$ ...

12

You are most likely getting an inaccurate value. As anticipated in Poutnik's comment above is difficult to reliably measure a $\mathrm{pH}$ of very acidic solution but the $\mathrm{pH}$ scale is indeed open, and negative $\mathrm{pH}$ are a real thing. It is easy and more meaningful to state a very high concentration or activity rather than reporting ...

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