Stoichiometry and Chemical Solutions Study Guide (page 2)
Homogeneous solutions are described by how much of a substance is dissolved in a specific solvent. The resulting concentration can be represented by many different units depending on the substance, the solvent, and the solution's potential use.
Concentration of Solutions
The concentration of a solution describes the amount of solute that is dissolved in a solvent. The solute is a substance that is dissolved in a liquid to form a solution (a homogeneous mixture). The liquid that the solute is dissolved in is the solvent. The concentration of a solution can be described in many different ways. The percent concentration expresses the concentration as a ratio of the solute's weight (or the volume) over the solution's weight (or the volume). This ratio is then multiplied by 100.
- Weight/volume % = grams of solute/100 mL of solvent
- Volume/volume % = volume of solute/100 volume of final solution
- Weight/weight % = grams of solute/100 g of solution
A 20% saline (NaCl or salt) solution contains 20 g of salt per 100 mL of water (20% = 20 g/100 mL * 100). Assuming they are homogeneously mixed, a 45% oil and vinegar solution would be 45 mL per 100 mL. In other words, 100 mL of the oil and vinegar solution would contain 45 mL of oil and 55 mL of vinegar. When dealing with two volumes, the total volume must be taken into consideration.
A more useful way to describe a concentration is molarity. Molarity (M) expresses the number of moles of solute per liter of solution. A 0.1 M NaOH aqueous solution has 0.1 mol of solute (NaOH) in 1 L of water. Because stoichiometric calculations require moles, molarity is more frequently used in calculations.
It is also useful to remember that grams can easily be converted to moles by using the molar mass of a substance.
What is the molarity of a 20% saline solution?
Molality (m) is the number of moles of a solute per kilogram of solvent.
What is the molality of a mixture made by dissolving 0.50 moles of sugar in 600 g water?
Normality (N) is the number of equivalents of the solute per liter of solution. A 1.0 N solution of acid (or base) contains one equivalent of an acid (or base) per liter of solution. A 1.0 M solution of HCl is 1.0 N, but a 1.0 M solution of H2SO4 is 2.0 N. Sulfuric acid has two acidic hydrogens, and the molarity is multiplied by a factor of 2. Phosphoric acid (H3PO4) is triprotic (having three protons it can donate) and a 1.0 molar solution is 3.0 normal.
The process of adding water to a solution is called dilution. Because only the solvent amount is changing, only the total volume and molarity of the solution is changing, not the number of moles of solute. Modifying the molarity equation yields
M1V1 = M2V2
The variables M1 and V1 represent the molarity and volume of the starting solutions, and M2 and V2 represent the final solution. Knowing three of the variables allows the algebraic calculation of the fourth variable.
What is the final molarity if 75 mL of 12 M concentrated HCl was diluted to 500 mL?
So, M1 = 12 M; V1 = 75 mL; M2 = ?; V2 = 500 mL
How much 3.0 M NaOH is needed to make 350 mL, 0.10 M NaOH solution?
So, M1 = 3.0 M; V1 = ?; M2 = 0.10 M; V2 = 350 mL
Part Per Million and Part Per Billion
A common gas and liquid concentration is parts per million and parts per billion. The "parts" measure the molecules per million or billion:
1 ppm = 1 molecule per every 1,000,000 molecules
1 ppb = 1 molecule per every 1,000,000,000 molecules
These concentration units are typically used when the concentrations are extremely small and it is impractical to use percent or molarity.
A titration involves the measurement of a titrant with a known concentration when reacted with an analyte (the substance being analyzed). The equivalence point or stoichiometric point of the reaction is where enough titrant has been added to exactly react with the analyte. The equivalence point of any base is the amount in grams that can be neutralized by 1 mole of H+ ions. The equivalence point of any acid is the amount in grams that can be neutralized by 1 mole of OH– ions. Usually, a substance called an indictor is added at the beginning to show a change by color in the equivalence point.
The equation M1V1 = M2V2 can be used to calculate the molarity of the analyte.
Exactly 45.23 mL of 19.2 M NaOH was titrated to the equivalence point with 75.0 mL of an unknown concentration of HCl. Calculate the molarity of the HCl solution.
Practice problems for these concepts can be found at - Stoichiometry and Chemical Solutions Practice Questions
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