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It depends how you write the reaction. For example, pure water + auto-ionized state, with some base added to remove some protons, will it auto-ionize a bit more (create more H+ and OH-) or a bit less (remove some H+ and OH-)? If we call the base $\ce{B}$ and the conjugate acid $\ce{BH+}$, you could write: $$\ce{B + H3O+ <=> BH+ + H2O}$$ This ...


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Assuming that by air bubble you're referring to the pocket of air now sealed in the bottle above the water surface: Immediately after sealing the bottle, the air pocket is not at all compressed: it's at the same pressure as the surrounding environment (the air outside the bottle). However, because you have a closed container, some water will evaporate, ...


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Making the solution in a single step is inaccurate up to impossible. Try to find a crystal of lead nitrate with the mass close enough to $\pu{1 mg}$ Try to weight up $\pu{1 mg}$ on scales with good enough accuracy. Try to make such $\pu{1 ppm}$ solution stable enough. The usual thing is to create a concentrated stock solution ( e.g $\pu{1000 ppm}$ and ...


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Russian interstate standard GOST 18293-72 "Drinking water. Methods for determination of lead, zinc and silver content" (PDF in Russian) suggests to use sulfarsazene (plumbone), which forms orange-yellow complex with lead(II). Reported method sensitivity (spectrophotometry): $\pu{0.5 μg L-1}.$ A brief review in English [1, p. 315]: To determine lead in ...


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You need to look for the data in a different way. You can use electrical conductivities to get the current flow the solution. From the imagined current flow you calculate how much hydrogen would be generated.


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$1$ liter water weighs $1030$ g, and contains $10.3$ mg $O_2$. The mass concentration of $O_2$ in ppm is the ratio $0.0103 g/1030$ g = $10.0$ ppm.


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The equilibrium reaction for the auto-dissociation of water is: $$2\text{ H}_2 \text{O}(l) \leftarrow \rightarrow \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$$ The associated equilibrium constant $K_w$ is: $$K_w=[\text{H}_3\text{O}^+]\times [\text{OH}^-] \approx 10^{-14}$$ (Strictly speaking the expression is: $$\frac{[\text{H}_3\text{O}^+]\times [\text{OH}^-...


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