How To Increase The Concentration Of A Solution
eight.1: Concentrations of Solutions
- Page ID
- 58828
Learning Outcomes
- Define concentration.
- Use the terms concentrated and dilute to describe the relative concentration of a solution.
- Calculate the molarity of a solution.
- Calculate percentage concentration (m/m, v/five, chiliad/5).
- Describe a solution whose concentration is in \(\text{ppm}\) or \(\text{ppb}\).
- Use concentration units in calculations.
- Determine equivalents for an ion.
- Complete calculations relating equivalents to moles, volumes, or mass.
- Complete dilution calculations.
In that location are several ways to express the amount of solute present in a solution. The concentration of a solution is a measure of the corporeality of solute that has been dissolved in a given amount of solvent or solution. A full-bodied solution is i that has a relatively large amount of dissolved solute. A dilute solution is i that has a relatively small-scale amount of dissolved solute. Even so, these terms are relative, and we need to be able to express concentration in a more exact, quantitative manner. Still, concentrated and dilute are useful as terms to compare one solution to another (see figure beneath). Also, be enlightened that the terms "concentrate" and "dilute" can be used as verbs. If you were to heat a solution, causing the solvent to evaporate, yous would be concentrating it, because the ratio of solute to solvent would be increasing. If you lot were to add together more than h2o to an aqueous solution, you would be diluting it because the ratio of solute to solvent would exist decreasing.
Per centum Concentration
1 mode to describe the concentration of a solution is by the percent of the solution that is composed of the solute. This percentage can exist determined in ane of three ways: (1) the mass of the solute divided by the mass of solution, (ii) the book of the solute divided by the volume of the solution, or (iii) the mass of the solute divided by the volume of the solution. Because these methods mostly issue in slightly different vales, it is important to ever indicate how a given percentage was calculated.
Mass Percent
When the solute in a solution is a solid, a convenient way to express the concentration is a mass per centum (mass/mass), which is the grams of solute per \(100 \: \text{g}\) of solution.
\[\text{Percentage past mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%\]
Suppose that a solution was prepared past dissolving \(25.0 \: \text{g}\) of sugar into \(100 \: \text{k}\) of h2o. The percent by mass would exist calculated equally follows:
\[\text{Percent by mass} = \frac{25 \: \text{g sugar}}{125 \: \text{g solution}} \times 100\% = 20\% \: \text{carbohydrate}\]
Sometimes, you may want to brand a particular amount of solution with a certain percent by mass and will need to calculate what mass of the solute is needed. For example, let'south say y'all demand to make \(three.00 \times 10^three \: \text{g}\) of a sodium chloride solution that is \(5.00\%\) past mass. You tin rearrange and solve for the mass of solute.
\[\begin{align} \% \: \text{by mass} &= \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\% \\ five.00\% &= \frac{\text{mass of solute}}{3.00 \times 10^3 \: \text{g solution}} \times 100\% \\ \text{mass of solute} &= 150. \: \text{g} \end{align}\]
You would need to counterbalance out \(150 \: \text{grand}\) of \(\ce{NaCl}\) and add together it to \(2850 \: \text{g}\) of water. Notice that it was necessary to subtract the mass of the \(\ce{NaCl}\) \(\left( 150 \: \text{grand} \correct)\) from the mass of solution \(\left( 3.00 \times x^3 \: \text{g} \right)\) to calculate the mass of the water that would need to be added.
Volume Per centum
The per centum of solute in a solution can more easily exist adamant past book when the solute and solvent are both liquids. The book of the solute divided by the book of the solution expressed every bit a percent, yields the percent by volume (book/volume) of the solution. If a solution is made by taking \(twoscore. \: \text{mL}\) of ethanol and adding enough water to make \(240. \: \text{mL}\) of solution, the percent by volume is:
\[\begin{align} \text{Percentage by volume} &= \frac{\text{volume of solute}}{\text{volume of solution}} \times 100\% \\ &= \frac{40 \: \text{mL ethanol}}{240 \: \text{mL solution}} \times 100\% \\ &= 16.7\% \: \text{ethanol} \end{align}\]
Ofttimes, ingredient labels on food products and medicines have amounts listed as percentages (encounter figure below).
It should be noted that, dissimilar in the case of mass, you cannot but add the volumes of solute and solvent to get the final solution volume. When calculation a solute and solvent together, mass is conserved, only volume is not. In the example above, a solution was made by starting with \(40 \: \text{mL}\) of ethanol and adding enough water to make \(240 \: \text{mL}\) of solution. Simply mixing \(twoscore \: \text{mL}\) of ethanol and \(200 \: \text{mL}\) of water would not give you the same result, as the final book would probably not be exactly \(240 \: \text{mL}\).
The mass-volume percent is also used in some cases and is calculated in a similar way to the previous two percentages. The mass/volume percent is calculated by dividing the mass of the solute past the book of the solution and expressing the result every bit a percent.
For example, if a solution is prepared from \(10 \: \ce{NaCl}\) in enough water to make a \(150 \: \text{mL}\) solution, the mass-book concentration is
\[\brainstorm{align} \text{Mass-book concentration} & \frac{\text{mass solute}}{\text{book solution}} \times 100\% \\ &= \frac{10 \: \text{g} \: \ce{NaCl}}{150 \: \text{mL solution}} \times 100\% \\ &= 6.seven\% \stop{marshal}\]
Parts per 1000000 and Parts per Billion
Two other concentration units are parts per million and parts per billion. These units are used for very small concentrations of solute such every bit the amount of lead in drinking water. Understanding these two units is much easier if you consider a percentage as parts per hundred. Recall that \(85\%\) is the equivalent of 85 out of a hundred. A solution that is \(xv \: \text{ppm}\) is fifteen parts solute per one million parts solution. A \(22 \: \text{ppb}\) solution is 22 parts solute per billion parts solution. While there are several ways of expressing two units of \(\text{ppm}\) and \(\text{ppb}\), we will treat them equally \(\text{mg}\) or \(\mu \text{g}\) of solutes per \(\text{L}\) solution, respectively.
For case, \(32 \: \text{ppm}\) could exist written as \(\frac{32 \: \text{mg solute}}{ane \: \text{L solution}}\) while \(59 \: \text{ppb}\) tin exist written every bit \(\frac{59 \: \mu \text{m solute}}{1 \: \text{L solution}}\).
Molarity
Chemists primarily need the concentration of solutions to be expressed in a way that accounts for the number of particles present that could react according to a particular chemical equation. Since pct measurements are based on either mass or volume, they are generally non useful for chemical reactions. A concentration unit based on moles is preferable. The molarity \(\left( \text{M} \correct)\) of a solution is the number of moles of solute dissolved in 1 liter of solution. To calculate the molarity of a solution, you lot carve up the moles of solute by the volume of the solution expressed in liters.
\[\text{Molarity} \: \left( \text{One thousand} \right) = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{\text{mol}}{\text{L}}\]
Note that the volume is in liters of solution and not liters of solvent. When a molarity is reported, the unit is the symbol \(\text{M}\), which is read as "molar". For example, a solution labeled as \(i.five \: \text{M} \: \ce{NH_3}\) is a "1.5 molar solution of ammonia".
Instance \(\PageIndex{1}\)
A solution is prepared past dissolving \(42.23 \: \text{one thousand}\) of \(\ce{NH_4Cl}\) into enough h2o to make \(500.0 \: \text{mL}\) of solution. Calculate its molarity.
Solution
Step ane: Listing the known quantities and program the trouble.
Known
- Mass of \(\ce{NH_4Cl} = 42.23 \: \text{chiliad}\)
- Molar mass of \(\ce{NH_4Cl} = 53.50 \: \text{g/mol}\)
- Book of solution \(= 500.0 \: \text{mL} = 0.5000 \: \text{L}\)
Unknown
- Molarity \(= ? \: \text{M}\)
The mass of the ammonium chloride is first converted to moles. Then, the molarity is calculated by dividing by liters. Notation that the given volume has been converted to liters.
Step two: Solve.
\[42.23 \: \text{g} \: \ce{NH_4Cl} \times \frac{i \: \text{mol} \: \ce{NH_4Cl}}{53.50 \: \text{1000} \: \ce{NH_4Cl}} = 0.7893 \: \text{mol} \: \ce{NH_4Cl}\]
\[\frac{0.7893 \: \text{mol} \: \ce{NH_4Cl}}{0.5000 \: \text{L}} = 1.579 \: \text{Grand}\]
Pace 3: Think virtually your outcome.
The molarity is \(1.579 \: \text{M}\), meaning that a liter of the solution would contain 1.579 moles of \(\ce{NH_4Cl}\). Having 4 pregnant figures is advisable.
Dilutions
When additional water is added to an aqueous solution, the concentration of that solution decreases. This is because the number of moles of the solute does not alter, but the total volume of the solution increases. We can ready an equality between the moles of the solute before the dilution (1) and the moles of the solute after the dilution (two).
\[\text{mol}_1 = \text{mol}_2\]
Since the moles of solute in a solution is equal to the molarity multiplied by the volume in liters, nosotros can set those equal.
\[M_1 \times L_1 = M_2 \times L_2\]
Finally, because the 2 sides of the equation are set equal to one another, the book tin can be in whatsoever units we choose, equally long as that unit of measurement is the same on both sides. Our equation for calculating the molarity of a diluted solution becomes:
\[M_1 \times V_1 = M_2 \times V_2\]
Additionally, the concentration tin be in any other unit as long every bit \(M_1\) and \(M_2\) are in the aforementioned unit of measurement.
Suppose that y'all have \(100. \: \text{mL}\) of a \(2.0 \: \text{G}\) solution of \(\ce{HCl}\). You lot dilute the solution by adding enough water to make the solution book \(500. \: \text{mL}\). The new molarity tin can easily be calculated by using the above equation and solving for \(M_2\).
\[M_2 = \frac{M_1 \times V_1}{V_2} = \frac{two.0 \: \text{K} \times 100. \: \text{mL}}{500. \: \text{mL}} = 0.40 \: \text{M} \: \ce{HCl}\]
The solution has been diluted by a gene of five, since the new book is 5 times as great as the original volume. Consequently, the molarity is one-5th of its original value. Some other common dilution problem involves deciding how much a highly full-bodied solution is required to brand a desired quantity of solution with a lower concentration. The highly concentrated solution is typically referred to as the stock solution.
Example \(\PageIndex{2}\)
Nitric acid \(\left( \ce{HNO_3} \correct)\) is a powerful and corrosive acid. When ordered from a chemical supply company, its molarity is \(sixteen \: \text{M}\). How much of the stock solution of nitric acrid needs to be used to make \(8.00 \: \text{L}\) of a \(0.l \: \text{M}\) solution?
Solution
Step i: List the known quantities and plan the problem.
Known
- Stock \(\ce{HNO_3} \: \left( M_1 \right) = 16 \: \text{Grand}\)
- \(V_2 = 8.00 \: \text{L}\)
- \(M_2 = 0.50 \: \text{Thousand}\)
Unknown
- Volume of stock \(\ce{HNO_3} \: \left( V_1 \right) = ? \: \text{L}\)
The unknown in the equation is \(V_1\), the necessary volume of the concentrated stock solution.
Step 2: Solve.
\[V_1 = \frac{M_2 \times V_2}{V_1} = \frac{0.l \: \text{M} \times 8.00 \: \text{L}}{sixteen \: \text{M}} = 0.25 \: \text{L} = 250 \: \text{mL}\]
Step 3: Think nigh your result.
\(250 \: \text{mL}\) of the stock \(\ce{HNO_3}\) solution needs to be diluted with water to a concluding book of \(8.00 \: \text{Fifty}\). The dilution from \(16 \: \text{M}\) to \(0.5 \: \text{Thousand}\) is a factor of 32.
Equivalents
Concentration is important in healthcare considering it is used in so many ways. Information technology'south also disquisitional to use units with any values to ensure the correct dosage of medications or study levels of substances in blood, to name just two.
Some other way of looking at concentration such as in Four solutions and blood is in terms of equivalents. One equivalent is equal to i mole of charge in an ion. The value of the equivalents is always positive regardless of the accuse. For example, \(\ce{Na^+}\) and \(\ce{Cl^-}\) both have ane equivalent per mole.
\[\begin{array}{ll} \textbf{Ion} & \textbf{Equivalents} \\ \ce{Na^+} & 1 \\ \ce{Mg^{2+}} & 2 \\ \ce{Al^{3+}} & 3 \\ \ce{Cl^-} & i \\ \ce{NO_3^-} & 1 \\ \ce{SO_4^{2-}} & 2 \stop{array}\]
Equivalents are used because the concentration of the charges is important than the identity of the solutes. For example, a standard Four solution does not contain the aforementioned solutes as blood merely the concentration of charges is the same.
Sometimes, the concentration is lower in which example milliequivalents \(\left( \text{mEq} \right)\) is a more than appropriate unit. Just like metric prefixes used with base units, milli is used to modify equivalents and then \(1 \: \text{Eq} = 1000 \: \text{mEq}\).
Example \(\PageIndex{iii}\)
How many equivalents of \(\ce{Ca^{two+}}\) are present in a solution that contains three.five moles of \(\ce{Ca^{ii+}}\)?
Solution
Use the human relationship between moles and equivalents of \(\ce{Ca^{2+}}\) to discover the answer.
\[3.5 \: \text{mol} \cdot \frac{2 \: \text{Eq}}{1 \: \text{mol} \: \ce{Ca^{two+}}} = seven.0 \: \text{Eq} \: \ce{Ca^{2+}}\]
Example \(\PageIndex{4}\)
A patient received \(1.50 \: \text{L}\) of saline solution which has a concentration of \(154 \: \text{mEq/L} \: \ce{Na^+}\). What mass of sodium did the patient receive?
Solution
Utilise dimensional assay to ready up the problem based on the values given in the problem, the human relationship for \(\ce{Na^+}\) and equivalents and the molar mass of sodium. Notation that if this trouble had a different ion with a different accuse, that would need to exist accounted for in the calculation.
\[1.50 \: \text{L} \cdot \frac{154 \: \text{mEq}}{1 \: \text{L}} \cdot \frac{1 \: \text{Eq}}{k \: \text{mEq}} \cdot \frac{1 \: \text{mol} \: \ce{Na^+}}{1 \: \text{Eq}} \cdot \frac{22.99 \: \text{g}}{one \: \text{mol} \: \ce{Na^+}} = 5.31 \: \text{one thousand} \: \ce{Na^+}\]
Contributors and Attributions
-
Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
Source: https://chem.libretexts.org/Courses/University_of_Kentucky/UK%3A_CHE_103_-_Chemistry_for_Allied_Health_(Soult)/Chapters/Chapter_8%3A_Properties_of_Solutions/8.1%3A_Concentrations_of_Solutions
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