# Tag Info

2

pH is not defined versus normality or molarity of an acid: Whatever the normality or the molarity, the pH is defined from the activity (or the concentration) of the ions H+. A given value of the normality or of the molarity does not give you the activity (or the concentration) of H+. So it does not allow you to calculate the pH.

4

Current definition implies that $\mathrm{pH}$ is a function of relative activity. Originally, the amount concentration of $\ce{H+}$ in $\pu{mol L-1}$ was proposed, which is also often used these days as an approximation . $\mathrm{pH}$ was originally defined by Sørensen in 1909 … in terms of the concentration of hydrogen ions (in modern nomenclature) ...

1

The value $\pu{1.3653 mol L^{−1} atm^{−1}}$ is the solubility constant (or Henry's law constant), not the solubility. The solubility is defined as the maximum concentration of a solute in solution under given conditions (temperature and pressure), whereas the solubility constant $K_\mathrm{H}$ defines how solute partitions between the gas (concentration in ...

1

Try adding pure salt to aqueous CuSO4 and freeze out Glauber's salt (Na2SO4.10H2O): 2 NaCl + CuSO4 (aq) --Freeze--> Na2SO4.10H2O + CuCl2 (aq) This path avoids the formation of problematic CaSO4 to quote a source (http://homepages.see.leeds.ac.uk/~earlgb/Publications/Vam%20Driessche%20et%20al%20New%20Perspectives%20on%20Mineral%20Nucleation%20and%20Growth%...

1

The problem could be solved with simultaneous equations, but the following are the wrong equations. $$K_{\mathrm{sp,}\ \ce{AgCl}} = [\ce{Ag+}][\ce{Cl-}]$$ $$K_{\mathrm{sp,}\ \ce{AgBr}} = [\ce{Ag+}][\ce{Br-}]$$ The right equations to use would be: $$K_{\mathrm{sp,}\ \ce{AgCl}} \ge [\ce{Ag+}][\ce{Cl-}]$$ $$K_{\mathrm{sp,}\ \ce{AgBr}} \ge [\ce{Ag+}][\ce{... 2 Both chloride and bromide ions are present in 10-fold excess over silver ions. That means that the chloride and bromide concentration in solution will not drop by much (they will remain major species). The solubility product of AgCl is 200-times higher than that of AgBr. If both AgCl and AgBr precipitate, the chloride solution would be 200-times higher than ... 4 You got the solubility part reversed. The solubility of \ce{AgCl} is lower than the solubility of \ce{Ag2CrO4}:$$s(\ce{AgCl}) = \sqrt{K_\mathrm{sp}(\ce{AgCl})} = \sqrt{\pu{1.8E-10 mol2 L-2}} = \pu{1.34E-5 mol L-1}s(\ce{Ag2CrO4}) = \sqrt{\frac{K_\mathrm{sp}(\ce{Ag2CrO4})}{4}} = \sqrt{\frac{\pu{1.1E-12 mol3 L-3}}{4}} = \pu{6.50E-5 mol L-1}$$... 0 A structured approach to this is to use at least five mixtures of the solvents and saturate with caffeine, stirring for at least four hours. The five mixtures are not picked arbitrarily but can be set at idealised levels. If you use 100% ethanol and water as the extremes, then you want 50:50 and two inbetween on either side. These can them be combined ... 1 Apparently you believe that \ce{NaOH} may be an independent molecule. It is not in this case. \ce{NaOH} does not exist in a solution. \ce{NaOH} does not exist in the solid state either. \ce{NaOH} does not exist in the solid state as a molecular compound. In the solid state, it is made of a huge pile of \ce{Na+} and \ce{OH-} ions, exactly like a ... 4 Based on the comment to an answer elsewhere by OP, I'd try to solve this problem, purely based on volumes. However, these calculations has ignored the volume contraction of 96% ethanol may have showed initially (that should be minimal since it is only ~4% of water by volume in there). I'd say, your equation, regardless of how OP has derived it, is erroneous.... 7 Measuring out volumes can sometimes be easier than dealing with mass, but in this case, converting the calculations to mass-centric can make the process easier. (I'll use your symbols, but they will now be in mass units (kg). If you desire percentages of ethanol by volume, you still need to convert to mass units because of the volume contraction!)) First, ... 2 A simple way to think about this is as follows (assuming we are talking about an aqueous solution). The dissolution of \ce{CO2} in water is followed by three chemical reactions:$$ \begin{align} \ce{CO2 + H2O &<=> H2CO3}\label{rxn:R1}\tag{R1}\\ \ce{H2CO3 &<=> HCO3- + H+}\label{rxn:R2}\tag{R2}\\ \ce{HCO3- &<=> CO3^2- + H+}\...

0

Yes, your voltage will change in accordance with the Nernst equation $$\Delta E_\text{cell} = - \frac{RT}{zF}\ln Q_{\text{rxn}}$$ So in your copper-zinc battery, where $$\ce{Cu^2+(aq) + Zn(s) -> Cu(s) + Zn^2+(aq)}$$ with a standard $E^\circ_\text{cell} \approx 1.10\text{V}$, setting the $\left[\ce{Cu^2+}\right] = 4\text{M}$ should raise your cell ...

7

I think that the cheapest and fastest way is that of working on natural products, like actual tea leaves, or ground coffee (I just tried it with coffee, and it can be done): If you want only particles with neutral buoyancy, or, if I understand well, floating in the solvent, you can use said natural products, which contain particles of different shapes and ...

2

Iron(III) chloride indeed hydrolyses in water, or to be precise, $\ce{Fe^3+}$ does assuring acidic medium. Hydrolysis isn't completely suppressed even in strongly acidic solutions. In the absence of other factors, what happens exactly depends on the concentration of chloride and temperature. In diluted cold solution a typical formation of metal aqua ions ...

1

Of course there are such solutions. Take, as an example, Sigma-Alrich's offer of methanolic solution of HCl (3 mol/L, e.g. here), or the dry ones in diethyl ether (e.g., 2 mol/L here), in cyclopentyl methyl ether (e.g., here), or in 1,4-dioxane (e.g. 4 mol/L, here).

2

Actually, I think the question has an error in the EMF of the cell given. It should be $\pu{-0.413 V}$ rather than $\pu{0.413 V}.$ So, actually you are right in saying that the left half cell would be cathode if the EMF of the cell has to be positive. The mistake that you are doing is when you wrote the reduction potential of anode, you should have written ...

8

If you dissolve $\ce{NH4Cl}$ in water, your solution contains $\ce{NH4+}$ and $\ce{Cl-}$ ions. If you dissolve $\ce{NaOH}$ in water, your solution contains $\ce{Na+}$ and $\ce{OH-}$ ions. Now if you mix those two solutions, nearly all $\ce{NH4+}$ and $\ce{OH-}$ will react and produce $\ce{NH3}$ and $\ce{H2O}.$ This shows that $\ce{NH4OH}$ cannot exist. It is ...

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