# Presentation: 002692

## Aqueous SolutionsConcentration / Calculations

*Dr. Ron Rusay*
*Spring 2002*

- © Copyright 1995-2002 R.J. Rusay

## Solutions

*Homogeneous solutions are comprised of **solute**(s), the substance(s) dissolved, [The lesser amount of the component(s) in the mixture], and*
*solvent**, the substance present in the largest amount.*
*Solutions with less solute dissolved than is physically possible are referred to as “**unsaturated**”. Those with a maximum amount of solute are “**saturated**”.*
*Occasionally there are extraordinary solutions that are “**supersaturated**” with more solute than normal.*

## Concentration and Temperature

*Relative Solution*
*Concentrations:*
***Saturated***
***Unsaturated***
***Supersaturated***

## DHMO, dihydromonoxide : “The Universal” Solvent

- _[http://www.](http://www.dhmo.org/)__[dhmo](http://www.dhmo.org/)__[.org](http://www.dhmo.org/)_

## Water : “The Universal” Solvent

- Generally, likes dissolve likes, i.e. polar-polar and nonpolar-nonpolar. Polar and nonpolar do not mix , eg. oil and water

- The oil (nonpolar) and water (polar) mixture don’t mix and are **immiscible**. If liquids form a homogeneous mixture, they are **miscible**.

## Aqueous Reactions & Solutions

*Many reactions are done in a homogeneous liquid or gas phase which generally improves reaction rates.*
*The prime medium for many inorganic reactions is water which serves as a solvent (the substance present in the larger amount), but does not  react itself.*
*The substance(s) dissolved in the solvent is (are) the solute(s). Together they comprise a solution. The reactants would be the solutes.*
*Reaction solutions typically have less solute dissolved than is possible  and are “unsaturated”.*

## Salt dissolving in a glass of water

## Solution Concentration

*A solution’s concentration is the measure of the amount of solute dissolved. *
*Concentration is expressed in several ways. One way is mass percent.*
*Mass % = Mass solute / [Mass solute + Mass solvent ] **	**	**	**	**	**	**	**	**	**x100*
*What is the mass % of 65.0 g of glucose dissolved in 135 g of water?*
*	**	**Mass % = **65.0 g / [65.0 + 135]g x 100*
*	**	**	**   **= 32.5 % *

## Solution Concentration

*Concentration is expressed more importantly as molarity ****(M)****.*
*	**Molarity (M) = Moles solute / Liter**	**	**	**	**	**	**	**Solution*
*An important relationship is ****M x V****solution****= mol***
*This relationship can be used directly in mass calculations of chemical reactions.*
*What is the molarity of a solution of 1.00 g KCl in 75.0 mL of solution?*
***M ******KCl ******=****** [1.00g ******KCl****** / 75.0mL ][1mol******KCl****** / 74.55 g ******KCl ******][ 1000mL / L]***
***	******= 0.18 mol******KCl****** / L***

## Solution Concentration

*The following formula can be used in dilution calculations:*
*	**	**	** ****M******1******V******1 ******= M******2******V******2***
*A concentrated stock solution is much easier to prepare and then dilute rather than preparing a dilute solution directly. Concentrated sulfuric acid is 18.0M. What volume would be needed to prepare 250.mL of a 1.50M solution?   *
***V******1 ******= M******2******V******2  ******/ M******1 ***
***V******1 ******= 1.50 M x 250. mL / 18.0 M ***
***V******1 ******= 20.8 mL***

## Solution Applications

*A solution of barium chloride was prepared by dissolving 26.0287 g  in water to make 500.00 mL of solution. What is the concentration of the barium chloride solution? ****M******BaCl2****** = ?***
***M******BaCl2****** *****=**
**=***** ******[26.0287g ******BaCl2****** / 500.00mL ][1mol ******BaCl2****** / 208.23g ******BaCl2 ******] [1000mL / L]***
***	***

***	******=  0.25000 mol / L***

## Solution Applications

*10.00 mL of this solution was diluted to make exactly 250.00 mL of solution which was then used to react with a solution of potassium sulfate. What is the concentration of the diluted solution. ****M******2****** = ?**** *
***M******BaCl2***** ****= *****M******1***
***M******2 ******= M******1******V******1  ******/ V******2***
***M******2 ******= 0.25000 M****** ****** ******x 10.00 mL****** ******/ 250.00 mL***
***M******2 ******= 0.010000 M***

## Solution Applications

*20.00** mL of the barium chloride solution required **15.50** mL of the potassium sulfate solution to react completely. ****M******K2SO4****** = ?**** *
*	**	**	**BaCl**2**(aq)** + K**2**SO**4**(aq)** ****  ?  +  ?*
*	**BaCl**2**(aq)** + K**2**SO**4** **(aq)** **** **2** KCl**(aq)** + BaSO**4** **(s)*
*?****M******K2SO4****** ***** ****= *****[M******BaCl2******x V******BaCl2****** / V******K2SO4****** ] [? ******mol******K2SO4****** / ? ******mol******BaCl2 ******]***
**	****	****	****0.010000 *****mol******BaCl2 ****** ******x *****0.02000 *****L******BaCl2****** x ******1 ******mol******K2SO4****** ***
***	******	******	******	****** ******L******BaCl2****** ******x****** ***0.01550 ***L******K2SO4****** ******x******  1 ******mol******BaCl2 ***

*?****M******K2SO4****** ***** ****=**

*?****M******K2SO4****** ***** ****= **0.01290 ***mol******K2SO4****** ****** ******/ L******K2SO4 *****= **0.01290 ***M******K2SO4****** ***

## ?gKCl  = 0.010000 molBaCl2 / LBaCl2 x 0.02000 LBaCl2  X  2 molKCl / 1 molBaCl2 X  
74.55 gKCl /molKCl

*How many grams of potassium chloride are produced?*
*BaCl**2**(aq)** + K**2**SO**4**(aq)** ****    ?  **	**+       ?** *
*BaCl**2**(aq)** + K**2**SO**4** **(aq)** **** **2** KCl**(aq)** + BaSO**4** **(s)*

**?g****KCl**** ***** = 0.010000 *****mol****BaCl2 ****/ L****BaCl2 ****x *****0.02000 *****L****BaCl2****  X  *****2***** ****mol****KCl**** / *****1***** ****mol****BaCl2 ****X  **
***74.55***** g****KCl ****/mol****KCl**
**	****	**

***1***

***= ******0.02982******  *****g****KCl**

## MolBaCl2 = MBaCl2x VBaCl2
= 0.10 molBaCl2 / LBaCl2 x 0.02000 LBaCl2
1 molBaCl2

= 2.0 x 10-3 

*If ****20.00**** mL of a ****0.10**** M solution of barium chloride was reacted with ****15.00**** mL of a ****0.20**** M solution of potassium sulfate, what would be the theoretical yield of barium sulfate?*
* **BaCl**2**(aq)** + K**2**SO**4** **(aq)** **** 2 KCl**(aq)** + BaSO**4** **(s) *

**Mol****BaCl2**** = M****BaCl2****x V****BaCl2**
**=***** ******0.10****** *****mol****BaCl2 ****/ L****BaCl2**** ****x *****0.02000****** *****L****BaCl2**
**1 mol****BaCl2**
**= *****2.0 x 10******-3***** **

**Mol****K2SO4****= M****K2SO4****x V****K2SO4**
**=***** ******0.20****** *****mol****K2SO4 ****/ L****K2SO4**** ****x *****0.01500****** *****L****K2SO4**
**1 mol****K2SO4**
**= *****3.0 x 10******-3***

***Which is the Limiting Reagent?***

***2.0 x 10******-3****** ***** *****< 3.0 x 10******-3****** ***

***2.0 x 10******-3****** *****mol ***is limiting *

## Must use the limiting reagent:	 
	0.10 molBaCl2  x 0.02000 LBaCl2  x  1 molBaSO4 x  233.39 g BaSO4
		LBaCl2 			  1 molBaCl2   	            molBaSO4

*If ****20.00**** mL of a ****0.10**** M solution of barium chloride was reacted with ****15.00**** mL of a ****0.20**** M solution of potassium sulfate, what would be the theoretical yield of barium sulfate?*
* **BaCl**2**(aq)** + K**2**SO**4** **(aq)** **** 2 KCl**(aq)** + BaSO**4** **(s) *

**Must use the limiting reagent:****	**** **
***	******0.10****** *****mol****BaCl2 **** ****x *****0.02000****** *****L****BaCl2  ****x****  ****1 mol****BaSO4 ****x  233.39 g ****BaSO4**
**	****	****L****BaCl2**** ****	****	****	****  1 mol****BaCl2   ****	****            ****mol****BaSO4**

- =

***= 0.47 g***