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# Tag Info

## New answers tagged ph

4

I suspect this is very likely a sentence adapted from of one of the methods from the Russian State Pharmacopoeia (RSP), which has numerous entries for the adverb «потенциометрически» (Eng. “potentiometrically”). For instance, there is a nearly identical match in the normative section for the preparation of acetate buffer solution, ОФС.1.3.0003.15-1.27 ...

1

In electrolysis, $Cu^{2+}$ is reduced in two steps, first to $Cu^+$ and later on to $Cu$. Of course $Cu^+$ is automatically disproportioned into $Cu$ and $Cu^{2+}$ in acidic conditions. But in neutral solutions, $Cu^+$ may react with $H_2O$, forming an unwanted red precipitate of $Cu_2O$. This will affect electroplating of copper.

3

There is high probability the desired properties of ascorbic acid are directly related to acid being acidic and in the free acid form, i.e. not in the form of the ascorbate. It is hard to advise, not being familiar with this part of the applied food chemistry, it the effect of ascorbic acid happens in the cold dough conditioning stage and/or in the stage of ...

1

The correct expressions are: $$\mathrm{pH} = \frac{1}{2}(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2})\tag{1}$$ $$\ce{[H+]} = \sqrt{\frac{K_1K_2[\ce{HA-}] + K_1K_\mathrm{w}}{K_1 + [\ce{HA-}]}}\tag{2}$$ Equation 2 is an exact expression (neglecting activities vs. concentrations), but Expression 1 is an approximation. To dervice the approximate ...

5

Since the equilibrium constant for the formation of the complex ion is very large, I assume that $\ce{[Ni(CN)4^{2-}] >> [Ni^{2+}]}$ From the comments: [comments:] That assumption is incorrect. The equilibrium constant seems large, but the exponents are high, so it is misleading. In fact, more of the cyanide is in the form of HCN than in complex ...

-2

I had a different approach to the problem which I have highlighted below Notes : Initially, I have considered the entire reagents to have reacted and then considered the equilibrium to set in I have considered two equilibria simultaneously, one involving the dissociation of the Nickel complex and also the hydrolysis of the Cyanide ion because it is a ...

1

The former equation assumes $$[\ce{H2A}]\simeq [\ce{A^2-}]$$ due reaction $$\ce{ 2 HA- <=> H2A + A^2-}$$ The is possible with 2 simplifying conditions: The concentration of oxonium resp. hydroxide ions originated from water dissociation is much lower than concentration of the basic resp. acidic ampholyte form. $$[\ce{H2A}] \gg \sqrt{K_\mathrm{w}}$$...

-1

My approach is the following. There is exactly enough $\ce{Ni^{2+}}$ ion and CN- ions for making up the ion $\ce{Ni(CN)4^2-}$. As the stability constant of this complex is huge, and HCN is an extremely weak acid. I would admit that the rare HCN molecules remaining in the solution have no effect on the pH of water, which must be 7.

0

If the water contains significant amount of bicarbonates, part of released carbon dioxide may escape and original bicarbonate is finally replace by the dihydrogen or hydrogen citrate. $$\ce{ HCO3- + H3Citr -> H2Citr- + H2O + CO2 ^}$$ $$\ce{2 HCO3- + H3Citr -> HCitr^2- + 3 H2O + 2 CO2 ^}$$ So, the final $\mathrm{pH}$ may not differ much from the ...

0

Oxygenated tap water is rich in transition metals including Fe and Mn ions. Citric is a source of H+ and a good chelate and can drive a redox reaction in the presence of oxygen and H+ proceeding as follows: $$4 Fe(2+)/Mn(2+) + O2 + 2 H+ --> 4 Fe(3+)/Mn(3+) + 2 OH-$$ There is also a likely metal redox couple equilibrium(s) that can be effective in ...

0

Maybe citric acid have had a negligible effect on the pH. Everybody knows that in contact with air, pure water is carbonated by CO2 from the atmosphere. And the pH may go down to 5.5. Later on, the water may loose its CO2, and the pH goes back to 7

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