Chemistry I Honors Acids Lesson #2 Behaviors and Reactions

A Comparison of Acids and Bases ACIDS Have a “sour” taste React with metals to form hydrogen gas React with carbonates to form carbon dioxide Have pH values less than 7 Are neutralized by bases Turn blue litmus red BASES Have a “bitter” taste React with some metals, but not many Have pH values above 7 Are neutralized by acids Solutions are “slippery” to the touch Turn red litmus blue

Reactions of Acids - Metals These reactions are essentially Replacement reactions (“Single Displacement”). The metal replaces the Hydrogen in the acid (metal replaces metal). Most common example is a familiar reaction: Mg + 2 HCl  MgCl 2 + H 2 (Note that the replaced Hydrogen becomes diatomic)

Reactions of Metals - Carbonates These reactions are more involved. They start as Exchange reactions, but then there is a simultaneous decomposition that ends up forming 3 products.

Consider HCl + Na 2 (CO 3 ) The reaction starts as an exchange reaction 2 HCl + Na 2 (CO 3 )  H 2 (CO 3 ) + 2 NaCl But then the H 2 (CO 3 ) decomposes into water and carbon dioxide. This will happen every time. 2 HCl + Na 2 (CO 3 )  H 2 O + CO NaCl

Neutralization This is an important reaction. It is essentially an exchange reaction, but because it involves an acid and a base, it is given its own “name”. In general terms, an acid and a base “neutralize” each other to form water and “a salt”. In this context, “salt” simply means an ionic compound.

Hydrochloric acid + Sodium Hydroxide This is a neutralization reaction. We will examine it as an Exchange reaction to get to the products. HCl + Na(OH)  H(OH) + NaCl Remember the pattern for Exchange reactions Remember that water can be represented as H(OH) The products are water and “salt”

Sulfuric acid + potassium hydroxide This is another acid – base neutralization. We will treat it as an Exchange reaction to get to the products. H 2 (SO 4 ) + 2 K(OH)  2 H(OH) + K 2 (SO 4 ) Note the products – water and an ionic compound (a “salt”)

Titration Titration is one of the classic laboratory activities in Chemistry classes. The objective is usually to determine the concentration of either the acid solution or the base solution.

Procedure In a titration, we cause a reaction between an acid and a base. We know the concentration of one of the solutions – we are trying to find the concentration of the other. In a typical titration, we start with a measured volume of the acid in a flask. Then we add several drops of an “indicator”. This is a chemical that changes color when the acid and the base have neutralized each other.

Continuing Now we add the base to the same flask. At first, there is no color change, but when enough base has been added to neutralize all of the acid molecules, the color change occurs. This is called the “endpoint” of the titration. We measure the volume of the base that has been added and then we can calculate the concentration of the “unknown” acid or base.

The Titration Equation This is the equation that is used in the calculations associated with titrations. V a M a = V b M b Where: V a represents the volume of the acid in mL M a represents the molarity of the acid V b represents the volume of the base in mL M b represents the molarity of the base

Calculations Any of the four variables can be calculated in a problem. All that happens is that the problem info provides you three of the four variables. You simply multiply and divide. Example: You have 30 mL of a hydrochloric acid solution, but its concentration is unknown. Several drops of phenolphthalein (an indicator) are added and then the acid is titrated with a 0.25 molar solution of sodium hydroxide. The endpoint (color change) occurs when 20 ml of the base have been added. What is the concentration of the acid?

Solving Remember that the equation used is: V a M a = V b M b From the info in the problem, we can substitute in like this: (30 mL) x M a = (20 mL) x (0.25 M) Then multiplying the numbers of the right and dividing by the 30 mL gives the answer: M a = M