{"id":519,"date":"2017-08-21T10:00:59","date_gmt":"2017-08-21T10:00:59","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/chemistry\/?page_id=519"},"modified":"2017-09-11T05:30:06","modified_gmt":"2017-09-11T05:30:06","slug":"solution-formation-and-concentrations","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/chemistry\/solution-formation-and-concentrations\/","title":{"rendered":"Solution Formation and Concentrations"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\">\n<p><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/gas-laws\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/gas-laws-and-solutions\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/solution-interactions\">Next Lesson \u27a1<\/a><\/p>\n<\/div>\n<p><!-- UPDATE NEXT\/PREVIOUS ABOVE --><\/p>\n<p><!-- CONTENT STARTS HERE --><\/p>\n<h1 id=\"title\">Solution Formation and Concentrations<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson we will examine how particles interact to form solutions and review the different ways in which\u00a0concentration is expressed.<\/p>\n<h4>Previously we covered&#8230;<\/h4>\n<ul>\n<li>Analyzing the basic assumptions of kinetic molecular theory<\/li>\n<li>Describing temperature in terms of the motion of molecules<\/li>\n<li>Applying Graham\u2019s, Dalton\u2019s, Charles\u2019s, Boyle\u2019s, the combined gas law, and the ideal gas laws<\/li>\n<\/ul>\n<section>\n<h3>What is a Solution?<\/h3>\n<p>Whether a substance is a solution or not depends on the characteristics of the particles involved in\u00a0the substance. It is important to be able to differentiate between a\u00a0<abbr title=\"A mixture that contains particles in excess of 1000 nm, which settle out quickly or can be filtered out \">suspension<\/abbr>, a\u00a0<abbr title=\"A mixture that contains particles between 1 and 1000 nm in size, which do not settle out and may resemble a solution.\">colloid<\/abbr>,\u00a0and a\u00a0<abbr title=\"A homogeneous mixture containing solute particles 1 nm or smaller in size \">solution<\/abbr>. Simple\u00a0tests can\u00a0be conducted to determine which substance it is. If particles are visible and settle over time or\u00a0can be filtered out, it is a suspension. A sample of water taken from the edge of a lake that has\u00a0been recently disturbed is a good example of a suspension. If the particles are not visible or the\u00a0mixture seems cloudy and does not settle over time, it may be a colloid or a solution. To determine\u00a0which it is, shine a light through the material. If the beam of light is invisible, it is a true solution. If you see the light beam reflecting off small particles, you are observing the\u00a0<abbr title=\"The scattering of light by solute particles in a colloid \">Tyndall Effect<\/abbr> and the\u00a0substance is a colloid.<\/p>\n<h3>How Does a Solution Form?<\/h3>\n<p><abbr title=\"A solution in which water is solvent\">Aqueous solutions<\/abbr> form when the forces between solute-solvent particles are greater than the forces between\u00a0solute-solute particles and solvent-solvent particles. In the case of sodium chloride dissolving in\u00a0water, the forces that hold sodium ions and chloride ions together are ionic bonds. The forces that\u00a0hold the solvent-solvent particles \u201ctogether\u201d are the\u00a0<abbr title=\"The attraction of hydrogen atoms in a molecular compound for highly electronegative atoms (nitrogen, oxygen, or fluorine) in another molecule\">hydrogen\u00a0bonds<\/abbr> between\u00a0water molecules. When sodium chloride is placed in water, the negative portion of the polar water\u00a0molecule is attracted to the positive sodium ion. When enough water molecules are attracted to the\u00a0sodium, they remove the sodium from the <abbr title=\"A rigid, repeating pattern found in a solid\">crystal\u00a0lattice<\/abbr>.\u00a0Likewise, the positive portion of the polar water molecule is attracted to the chloride ion and when\u00a0there are enough water molecules surrounding the chloride ion, it is removed from the crystal\u00a0lattice.<\/p>\n<h3>What Terms are Used to Describe a Solution?<\/h3>\n<p>Solutions can be described with very general terms such as\u00a0<abbr title=\"A solution in which the maximum amount of solute is dissolved in a specific amount of solvent at a given temperature\">saturated<\/abbr>,\u00a0<abbr title=\"A solution in which less than the maximum amount of solute is dissolved in a specific amount of solvent at a given temperature\">unsaturated<\/abbr>,\u00a0and\u00a0<abbr title=\"A solution in which more than the maximum amount of solute is dissolved in a specific amount of solvent at a given temperature; this is generally achieved by heating the solvent and adding more solute, then cooling the solution gradually\">supersaturated<\/abbr>.\u00a0Often these terms relate to the amount of solute found in a specific amount of solvent on a solubility curve.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/solutionformationconcentration2.solubility_graph.gif\" alt=\"Solubility graph\" width=\"388\" height=\"471\" \/><\/center><\/p>\n<p>At a given temperature, when the amount of solute dissolved in the given amount of solvent falls below the line for\u00a0the substance, the solution is unsaturated. When the amount of solute dissolved is greater than the value shown by\u00a0the line, the solution is supersaturated. If the amount of solute dissolved is equal to the value shown by the line,\u00a0the solution is saturated.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Based on the graph above, a solution containing 45.0 g of KNO<sub>3<\/sub> in 200.0 cm<sup>3<\/sup> of water at\u00a030\u00baC would be___________________.<\/p>\n<ol>\n<li>saturated<\/li>\n<li>supersaturated<\/li>\n<li>unsaturated<\/li>\n<li>the answer cannot be determined from the information given<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is C. The graph shows that at 30 \u00baC, 45.0 g of KNO<sub>3<\/sub> would be the maximum amount of\u00a0solute that could be dissolved in 100.0 cm<sup>3<\/sup> of water. Based on this information, if you had 200.0 cm<sup>3\u00a0<\/sup>of water, 90.0 g of KNO<sub>3<\/sub> could be dissolved, which would be twice the amount given in the question.<\/p>\n<\/section>\n<h3>How Much is in a Solution?<\/h3>\n<p>Concentration is the term used to describe the amount of solute present in a solution. Concentration can be expressed\u00a0in a variety of units. When describing very small amounts of a solute in a large amount of solvent, parts per\u00a0million, or ppm may be used.<\/p>\n<h4>Volume\/volume%<\/h4>\n<p>Volume\/volume% is a unit used when both the solute and the solution can be measured in milliliters. In order to get a\u00a060% alcohol in water solution you would measure out 60 mL of alcohol and dilute it with water until your total\u00a0volume was 100 mL.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s2_003.gif\" width=\"198\" height=\"44\" \/><\/center><\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>What is the volume\/volume concentration of a solution that contains 65.0 mL of rubbing alcohol in 255 mL of aqueous\u00a0solution?<\/p>\n<ol>\n<li>0.203 %<\/li>\n<li>0.255 %<\/li>\n<li>20.3 %<\/li>\n<li>25.5 %<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is D. By using the formula <img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s2_003.gif\" width=\"198\" height=\"44\" align=\"absmiddle\" \/> and substituting the values in\u00a0properly we get:<\/p>\n<p class=\"q-reveal center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s2_005.gif\" width=\"229\" height=\"44\" \/><\/p>\n<p class=\"q-reveal\">The answer is 25.5 %. If you chose answer C, it is most likely because you added the two volumes (65 mL and 255 mL) to get a larger\u00a0solution volume. If you selected choice B, you just forgot to multiply by 100 to get the\u00a0percent. If you picked choice A, you probably used the larger volume and forgot to multiply by\u00a0100 to get the percent.<\/p>\n<\/section>\n<h4>Weight\/weight%<\/h4>\n<p>Weight\/weight% is a unit used when the weight of the solute can be divided by the weight of the solution and\u00a0multiplied by 100 to calculate the concentration of the solution.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s2_006.gif\" width=\"185\" height=\"44\" \/><\/center><\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>How much sodium chloride would be necessary to prepare 491 g of a solution whose <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s2_007.gif\" width=\"72\" height=\"44\" \/>concentration is 37.1 %?<\/p>\n<ol>\n<li>7.56 g<\/li>\n<li>37.1 g<\/li>\n<li>182 g<\/li>\n<li>491 g<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is C. By using the formula <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s2_006.gif\" width=\"185\" height=\"44\" \/> and substituting the values in\u00a0properly we get:<\/p>\n<p class=\"q-reveal center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s2_009.gif\" width=\"228\" height=\"44\" \/><\/p>\n<p class=\"q-reveal\">Solving for the g of solute, the result is 182 g.<\/p>\n<\/section>\n<h4>Molarity<\/h4>\n<p>The most commonly used unit of concentration in chemistry is molarity, <em>M<\/em>, which utilizes the number of moles\u00a0of solute relative to the volume of the solution.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s2_010.gif\" width=\"146\" height=\"44\" \/><\/center><\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>How many grams of sodium sulfate are needed to prepare 500 mL of 0.100 M solution?<\/p>\n<ol>\n<li>2.00 x 10<\/li>\n<li>0.0500 g<\/li>\n<li>5.95 g<\/li>\n<li>7.10 g<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is D. First, convert the volume from mL to L by dividing by 1000. Next, solve\u00a0for the number of moles of solute needed by substituting into the equation <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s2_010.gif\" width=\"146\" height=\"44\" \/> to\u00a0get <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s2_012.gif\" width=\"216\" height=\"44\" \/><br \/>\nThe number of moles necessary will equal 0.0500 moles. To get the number of grams of sodium sulfate,\u00a0multiply by the molar mass of sodium sulfate. Remember the formula for sodium sulfate is\u00a0Na<sub>2<\/sub>SO<sub>4<\/sub>, so the molar mass is 142.00 g\/mole (2<strong>\u2219<\/strong>22.99 + 32.06 +\u00a04<strong>\u2219<\/strong>15.99).<\/p>\n<\/section>\n<h3>Dilution Problems<\/h3>\n<p>Based on molar concentration, or molarity, a simple formula can be employed to determine the concentration of a\u00a0solution that has been diluted from one of a stronger concentration, sometimes referred to as a stock solution.<\/p>\n<p class=\"center\" style=\"text-align: center;\"><em>M<\/em><sub>1<\/sub><em>V<\/em><sub>1<\/sub>=<em>M<\/em><sub>2<\/sub><em>V<\/em><sub>2<\/sub><\/p>\n<p><em>M<\/em><sub>1<\/sub> is the concentration of the stock solution and <em>V<\/em><sub>1<\/sub> is the volume that you\u00a0measure to be diluted. <em>M<\/em><sub>2<\/sub> is the concentration of the dilute solution and <em>V<\/em><sub>2\u00a0<\/sub>is the volume of your final solution.<\/p>\n<h3>Molality<\/h3>\n<p>Another concentration unit used is molality, <em>m<\/em>, which compares the number of moles of solute to the mass of\u00a0the solvent. We will revisit this unit of concentration later when we discuss colligative properties.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s2_015.gif\" width=\"145\" height=\"44\" \/><\/center><\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>How many mL of concentrated hydrochloric acid (HCl) would be necessary to prepare 250.0 mL of 3.00 M HCl?<br \/>\nConcentrated HCl is 12.0 M.<\/p>\n<ol>\n<li>62.5 mL<\/li>\n<li>144 mL<\/li>\n<li>188 mL<\/li>\n<li>1.00 \u00d7 10<sup>3<\/sup> mL<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is A. Sometimes it is helpful to make a table of the given information to solve a problem.<\/p>\n<p class=\"q-reveal center\">\n<table class=\"q-reveal\" cellspacing=\"0\" cellpadding=\"0\">\n<thead>\n<tr>\n<td valign=\"top\" width=\"153\">Initial<\/td>\n<td valign=\"top\" width=\"153\">Final<\/td>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td valign=\"top\" width=\"153\"><em>M<\/em><sub>1<\/sub> = 12.0 M<\/td>\n<td valign=\"top\" width=\"153\"><em>M<\/em><sub>2<\/sub> = 3.00 M<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\" width=\"153\"><em>V<\/em><sub>1<\/sub> = ?<\/td>\n<td valign=\"top\" width=\"153\"><em>V<\/em><sub>2<\/sub>= 250.0 mL<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p class=\"q-reveal\">Using the\u00a0formula <em>M<\/em><sub>1<\/sub><em>V<\/em><sub>1<\/sub>=<em>M<\/em><sub>2<\/sub><em>V<\/em><sub>2<\/sub>,\u00a0you can solve for the first volume of acid by dividing both sides by <em>M<\/em><sub>1<\/sub>.<\/p>\n<\/section>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<p><!-- UPDATE NEXT\/PREVIOUS BELOW --><\/p>\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/gas-laws\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/gas-laws-and-solutions\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/solution-interactions\">Next Lesson\u00a0\u27a1<\/a><\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/section>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous Lesson\u00a0Workshop Index\u00a0Next Lesson \u27a1 Solution Formation and Concentrations Objective In this lesson we will examine how particles interact to form solutions and review the different ways in which\u00a0concentration is expressed. Previously we covered&#8230; Analyzing the basic assumptions of kinetic molecular theory Describing temperature in terms of the motion of molecules Applying Graham\u2019s, Dalton\u2019s, [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-519","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/519","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/comments?post=519"}],"version-history":[{"count":13,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/519\/revisions"}],"predecessor-version":[{"id":729,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/519\/revisions\/729"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/media?parent=519"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}