# Tag Info

39

You are correct suggesting that 1 μg/kg implies 1 ppb, however the reverse is not true. For instance, 1 ppb can also be 1 nmol/mol, and the reader will never have a chance to deduce which one is it unless you explicitly define the usage of the "parts per something" in the text. This clutters the manuscript with redundant notes and causes overall confusion. ...

34

It depends how you dilute it. If you take an aqueous solution of A and just add pure water (absolutely 100% water), the concentration of A will never quite be null. In this case however, you will reach a point where the concentration of A is so small that it can be considered null for your applications. If, however, you dilute the solution, take a sample, ...

19

One of the key attractions of molality is that changing the temperature of a solution does not change the molality, while it may change the molarity. This is because the volume of the solution changes as a result of expansion or contraction of the solvent upon changing the temperature, and thus the molarity changes, since $$M=\frac{n}{V}.$$ The downside, of ...

18

You correctly point out that the number of molecules in a solution is finite and constant, however the volumetric concentration (that is, how many molecules per litre) changes upon dilution. If, for instance, you take one liter of a 1 mol/L solution of ethanol in water ($\approx{6.02\times{10}^{23}}$ ethanol molecules per liter) and add 9 liters of distilled ...

16

Does the activity of a solid or liquid change over the course of a reaction? The density of a solid or liquid reactant doesn't change over the course of a reaction. The mass and volume do as it is consumed, but the ratio of the two is constant. If the reaction causes a temperature change then there are small changes in density, but that would also alter the ...

15

The textbook is precisely correct. The equilibrium constant $K$ which the logarithm is taken of is dimensionless, and includes activities or fugacities, and not concentrations and pressures. In practice this is achieved by using standard states which refer to the pure materials: standard concentration $c^⦵$ and standard pressure $p^⦵$. One must be very ...

14

In general, salt (particularly NaCl) will increase the rate of corrosion (rusting). To understand why, consider metallic iron $\ce{Fe}$ which rusts (oxidises) to iron(II) oxide $\ce{Fe2O3}$ in the presence of oxygen $\ce{O2}$ and water $\ce{H2O}$. $$\ce{4Fe +3O2 +6H2O->4Fe(OH)3}$$ Corrosion (rust) is a 'redox' reaction, which means it involves ...

14

The curly brackets denote "activity of" the species therein. See the Wikipedia section: Basic definitions and properties of Equilibrium constant

13

The term "active mass" is a historical term. The concept of an equilibrium constant was developed by Cato Maximilian Guldberg and Peter Waage. The Law of Mass Action has also been referred to as the Law of Guldberg and Waage, historically. Guldberg and Waage defined the term "active mass" in the 1867 Études sur les affinitès chimiques. l'on peut ...

13

Add Calcium Chloride to destroy the azeotrope. Then distill and capture the vapor phase. See the second method "Extractive Distillation" here: http://www.jacobs.com/uploadedFiles/wwwjacobscom/20_Learn_About_Us/25_Products/252_Chemetics/Hydrochloric%20Acid%20Concentration.pdf (Obviously, proper equipment and safety precautions are essential.)

13

This is a so-called "Pearson's square" or "Box method" of balancing ratios, originally used extensively in dairy industry (at least since 1920s judging from Google Books search). Earlier the similar approach has been used in sugar industry by using "Cobenz diagrams" aka spider diagrams. Widely popularized in Soviet books for analytical chemistry at least ...

13

For an analytical chemist, the concept of zero concentration does not exist. The concentration cannot be exactly 0! Only a limit of detection can be developed in terms of statistics. This is why a senior respectable user here has written an entire monograph on this topic. Suppose you have a NaCl solution, and so called "pure water"*, matter how ...

12

In addition to JSK's excellent answer, I'd like to point out that there's a common pitfall related to molarity (which JSKs answer might have slipped on): molarity is defined as amount of substance of solute per liter of solution, while molality is amount of substance of solute per kilogram of solvent. This might not seem like a huge difference, but if you ...

12

Chemical equilibrium is a type of dynamic equilibrium, but not every dynamic equilibrium is a chemical equilibrium. In a chemical equilibrium there is no change on the macroscopic scale. That means that if you look at the system it seems like nothing is happening, but at molecular scale there are reactions going on and the rate of forward reaction = rate of ...

12

A chemical equilibrium concerns chemical reactions. There should be at least a forward- and backward reaction between two species but more complex systems with multiple individual reactions may occur. The important observation is that there is no macroscopic change to the chemical constituents of the system, i.e. the concentrations of all reaction partners ...

11

What you witnessed was probably something similar to superheating/bumping. Essentially, when a solution is heated too rapidly, large pockets of vapor can form beneath the surface of the liquid that have an internal temperature higher than the expected boiling temperature of the liquid. In those situations, even small physical disturbances and/or further ...

11

Your problem is that you've assumed the density is 1 gram/mL. Remember that a molar is defined as a mole of solvent per liter of solution, not solvent. Usually, in introductory chemistry classes, we skip over the fact that adding solute to a solution increases its density, because it makes life more complicated. As you just found out though, sometimes you ...

10

$$K_\mathrm{a}=\ce{\frac{[A-][H3O+]}{[HA]}}$$ The above equation defines the $K_\mathrm{a}$, or acid dissociation constant, of an acid. The reason it is even a value worth measuring is because it is an intrinsic property of the acid, i.e. it's value does not change from sample to sample (temperature aside). While it is true that as a solution of acid ...

10

Irrespective of the potential hazards to an inexperienced chemist attempting distillation of hydrochloric acid, especially absence of adequate safety measures such as a fume hood, in this case, distillation will not result in a higher concentration of acid. Hydrogen chloride and water form a constant-boiling mixture (azeotrope) at ~20% HCl, which is what you ...

9

Molarity of diluted $\ce {H2SO4}$ (solution 2): $\pu{0.0013798 mol}/\pu{0.025 L} = \pu{0.054172 M}$ (I may be using the wrong volume, is it possible that I have to add the $\pu{25 ml}$ to the $\pu{23.81 ml}$ and divide by $\pu{0.04881 L}$?) No, you used the correct volume since you want to know the concentration of the $\pu{25 mL}$ that you added to the ...

9

Concentration is any measurement of the quantity of a solute that is present per unit of solution (in general, there are exceptions like moLality, which measures the number of moles of solute per kilogram of solvent). Molarity is just one way to express concentrations, which, more specifically, is the number of moles of solute per liter of solution. But ...

9

The user ssavec prompted me to research further on my own and post an answer. It seems the limits are practical ones. For hydrochloric acid — $\ce{HCl}$ a practical limit is 38% with absolute limit around $40~\%$ and commercial concentrations ranging from 30 to 35% for convenience in transportation infrastructure. For nitric acid — $\ce{HNO3}$ the $68~\%$ ...

8

Take $\pu{200 mL}$ of the $30\%$ solution, add water to a final volume of $1$ liter. Then you have a $6\%$ solution. Next, take $\pu{59 mL}$ of the $6\%$ solution and add water until it is $1$ liter. After the first dilution it's not exactly $6\%$, because the density of $30\%$ $\ce{H2O2}$ is $\pu{1.135 g/cm3}$. It's more like $6.6\%$, but the instructions ...

8

Yes. $$n = \frac{m}{M}~~~~~~~~~~n = cV$$ So $$c = \frac{m}{VM}$$ Given $\frac{m}{V}$ you can work out $c$ (in $\mathrm{mol~dm^{-3}}$).

8

Unequivocally yes! If your solute is something of which you can identify a single molecule in a macroscopic solvent sample (let´s say a fluorescent dye), then you can dilute down to zero, and be sure about it. (For the nitpickers: This can be done in a finite number of steps, if you split the solution in two after each step, and keep diluting the part which ...

7

The short answer is because $\ce{HF}$ doesn't dissociate as much as $\ce{HCl}$. Let's examine why this is by introducing some terms like $\ce{pH}$ and $\ce{pK_a}$. $\ce{pH}$ does not measure the strength of a given acid, but rather the acidity of a given solution. $\ce{pK_a}$ does measure the relative strength of an acid, look at the following ...

7

Your concentration is 0.615 M. The M stands for molar concentration, which is the number of moles per liter of solution. Thus, your concentration of 0.615 M (mol/L) means that in one liter of solution there will be 0.615 moles of solute (KCl). Similarly, a solution with concentration of 0.615 g/L would have 0.615 grams of solute per 1 liter of solution. A ...

7

You're using $2.5\ \mathrm g$ of solute out of $100\ \mathrm g$ of solution, which is $2.5\ \%$, but the initial question is asking about a $0.025\ \%$ solution, which is why your answer is two orders of magnitude too large. You should be using $0.025\ \mathrm g$ of solute and $99.975\ \mathrm g$ of solvent. As a general helpful hint for answering multiple ...

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