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In the $\pu{500 cm^3}$ solution there are is already $\ce{F^-}$ dissolved coming from the $\ce{NaF}$ $\pu{0.10 mol dm^{-3}}$ that was already there. This means that the solubility of $\ce{CaF2}$ is reduced due to the common ion effect. NB $\ce{NaF}$ can be considered completely dissociated because it has a very high $K_\mathrm{sp};$ therefore the initial ...


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Gc3941d has mixed the solubilities of $\ce{NaF}$ and $\ce{CaF_2}$. Let's start the calculation from the beginning. The concentration of fluoride ion is : [$\ce{F^-}$] = $0.1$ M. The concentration of Calcium may be calculated from the solubility product $\ce{K_{sp}}$ and [$\ce{F^-}$] according to : :$$\ce{[Ca^{2+}] = K_{sp} /[F^-]^2} = 1.46·10^{-10}/(0.1)^2 ...


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Answering this question requires a preliminary discussion of the solubility product constants of three silver halides and the formation constants of three silver complexes. First, note that $\ce{AgCl}$, $\ce{AgBr}$, and $\ce{AgI}$ are all insoluble in water, but insoluble is a relative term in the end. Their respective solubility equilibria and solubility ...


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Mercury cyanide behaves like Mercury chloride $\ce{HgCl_2}$. Both are soluble in water, and both do not dissociate in water. These compounds are not salts. The bonds $\ce{Hg-Cl}$ or $\ce{Hg-CN}$ are more covalent than ionic. When the atom Hg is included in compounds it often behaves as a non-metal. This is due to a relativistic effect. In all elements ...


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@Maurice - Your last paragraph is inconsistent with $K_\mathrm{sp} = 3.0\cdot10^{-17}$. You gave: $$\ce{[Zn^{2+}] = 1.95\cdot10^{-6}}$$ $$\ce{[Zn(OH)+] = 2.25\cdot10^{-6}}$$ Neglecting the autopyrolysis of water which is really an insignificant correction... $$\ce{[OH-] = (2\times1.95 + 2.25)\cdot10^{-6}} = 6.15\cdot10^{-6}$$ $$\therefore K_\mathrm{sp} =...


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There are no simple relation between the solubility and the solubility product when a doubly charged ion is involved. The measured solubility is always much bigger than the value obtained from the solubility product. This is due to the doubly charged ion. Doubly charged ions like $\ce{Zn^{2+}}$ are usually hydrolyzed in water, and are partly transformed ...


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The molar mass of zinc hydroxide is 99.424, so all your math checks out. Wikipedia also notes for zinc hydroxide that the $K_\mathrm{sp} = 3.0\times 10^{-17}$, so there is an inconsistency between the $K_\mathrm{sp}$ and the solubility data. Based on the $K_\mathrm{sp}$, the solubility should be $\pu{1.9\times10^{-4} g/L}$. Obviously one source of the ...


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So that must imply that with increased vapor pressure, the mole fraction of the gas in the solution must increase, but quite the opposite is true. Vapor pressure, or saturation pressure, $P_{sat}$, is different than system pressure, $P_{sys}$. $P_{sat}$ is an intensive property, relative to the compound of interest, while $P_{sys}$ is an extensive property ...


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Solubility of calcium sulphate is quite high, compared to many much less soluble salts, like barium sulphate, calcium phosphate or calcium fluoride. Most of insoluble minerals are salts. Insoluble/limited solubility salts have as crystals lower Gibbs energy than dissolved, leading to solution being thermodynamically unfavourable. That is related to ...


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All carbonated drinks are oversaturated solutions of $\ce{CO2(g)}$. Pouring over fruits lead to contact with many residual air bubbles, which serve as fizzing places for $\ce{CO2(aq) -> CO2(g)}$. It is more or less the same as the ( previously already opened ) drink shaking, where violent fizzing occurs as well. It is a similar process activation, ...


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Google is your friend. Using "Calcium chloride solubility", you find plenty of references where this solubility is given at different temperatures. Expressed in grams $\ce{CaCl2}$ in $100$ mL water, it is $59.5$ à $0°$C, $64.7$ at $10°$C, $100$ at $20°$C, $128$ at $30°$C, $137$ at $50°$C, $147$ at $70°$C, and $159$ at $90°$C. Do the same with Magnesium ...


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Raoult's law is a proposition regarding the relationship between the vapor pressure of a solvent (component 1 below) in its pure form ($p_1^\circ$) and when a solute (component 2) is present at a mole fraction $\chi_2=1-\chi_1$ (the vapor pressure then $p_1$): $$\chi_1 = \frac{p_1}{p_1^\circ} $$ It is generally true for very dilute solutions because we ...


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This graph is a very good example of experimental errors which may give rise to a pseudo-phenomenon i.e., an observation which apparently looks like a scientific phenomenon but it is an artifact of the experimental design. First of all work out the moles of acetic acid (5% acetic acid) and moles of NaHCO3. As you can see vinegar has become a limiting agent. ...


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Gist of the matter... When an ionic salt dissolves in water the ions and water molecules don't just bounce around independently. The ions have a charge and water molecules are polar. So the water molecules form a cluster, which is a number of concentric layers of water molecules, around the ions. A gas molecule dissolved in the water can't get into a ...


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This is all rust which has precipitated on the walls and it is very hard to remove it. You might have to use oxalic acid to remove rust stains on glass. Oxalic acid plays a dual role, it dissolves the rust like an acid, but it also complexes the iron, which prevents further hydrolysis and re-settling.


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In chemistry the Principle of Atom Conservation says that there must be the same number of like atoms on both sides of the equation. Balance the following chemical reaction by determining the appropriate values for $a, b, c, d$ and $e$ using Diophantine linear equations: $$\ce{aS + bHNO3 <=> cH2SO4 + dNO2 + eH2O}$$ $$\text{i.e. Sulfur + Nitric acid}\...


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Copper, $\ce{Cu}$, is oxidized by concentrated nitric acid, $\ce{HNO3}$, to produce $\ce{Cu^{2+}}$ ions. The nitric acid is reduced to nitrogen dioxide, $\ce{NO2}$, or nitric oxide, $\ce{NO}$, according to the following chemical equations. $$\ce{Cu + 4HNO3 -> Cu(NO3)2 + 2NO2(gas) + 2H2O}\tag{1}$$ i.e. When 1 atom of copper is dissolved then 2 ...


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How about this? Acetic acid, $\ce{CH3COOH}$, ionizes in water according to the following chemical equation to produce a hydronium ion, $\ce{H3O+}$, and an acetate anion, $\ce{CH3COO-}$. $$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$ The mathematical equation for the equilibrium is: $$1.8\cdot10^{-5} =\dfrac{\ce{[H3O+][CH3COO-]}}{\ce{[CH3COOH]}} $$ ...


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You may also give the weight of a sample made of an alloy $\ce{Zn-Mg}$. Then you dip it into some diluted $\ce{HCl}$. Both metals will react simultaneously according to $$\ce{Zn + 2 HCl -> H_2 + ZnCl_2}$$ and $$\ce{Mg + 2 HCl -> H_2 + MgCl_2}$$ This will produce an important amount of $\ce{H_2}$ gas. You can measure its volume. So you have ...


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I think a useful problem with the properties you describe is finding how a acid with several ionized forms changes with pH in the medium. Recently I explained how to do it for $\ce{CO2}$ here. Look How to calculate bicarbonate and carbonate from total alkalinity. The general solution scales nicely with complexity of the system: ACID TYPE DIFFERENT ...


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Your precipitate is very likely iron oxides. First make sure you get the remains of the hydrogen peroxide out completely (e.g. by flushing several times with a little distilled water), then use an acid to dissolve the oxide. Possibly acetic acid (vinegar) or citric acid and a little heating will do the job. If not, try 20% sulphuric acid or 10% hydrochloric ...


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The key concept here is that for solubility, you need to compare alcohols with the same number of carbon atoms. For example, the solubility order of isomers of butanol is as follows: $$\ce{CH3CH2CH2CH2OH < (CH3)2CHCH2OH < (CH3)3COH}$$ This difference in solubility follows the order 3°>2°>1° and can be explained by the branching concept that you have ...


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Ok, after having poked around a bit I doubt that $\ce{Pb(CN)2}$ forms. I'm guessing that you're collecting the froth to get a concentrate of the galena $\ce{PbS}$ out of some finely crushed ore. An aspect of the operation seems to be chemically reducing the concentration of $\ce{ZnS}$ by dissolving via the following reaction: $$\ce{ZnS(s) + 4NaCN(l) -&...


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