Net Ionic Equations Titration Buffers Ch 19 Acids

Net Ionic Equations, Titration, & Buffers Ch. 19 Acids & Bases

Strong Acid + Strong Base Instead of an overall equation, substances that primarily exist as ions can be written.

Spectator Ions • Na+ and Cl- don’t participate in the reaction, they are spectator ions. • Na+ & Cl- show up on both sides of the equation.

Net Ionic Equations • When spectator ions are removed from both sides of the equation, you have the net ionic equation.

Practice Net Ionic Equations • Write the net ionic equation for sulfuric acid, H 2 SO 4, and strontium hydroxide, Sr(OH)2.

Neutralization • Chemical reaction between an acid and a base. • Products are a salt (ionic compound) and water.

Neutralization ACID + BASE SALT + WATER HCl + Na. OH Na. Cl + H 2 O strong neutral HC 2 H 3 O 2 + Na. OH Na. C 2 H 3 O 2 + H 2 O weak strong basic • Salts can be neutral, acidic, or basic. • Neutralization does not mean p. H = 7.

Buffers • Resist changes in p. H when moderate amounts of acids or bases are added to the solution. • It contains ions or molecules that react with OH– or H+ if one of these ions is introduced into the solution • Buffer solutions are prepared by using a weak acid with one of its salts or a weak base with one of its salts

Buffers • For example, a buffer solution can be prepared by using the weak base ammonia, NH 3, and an ammonium salt, such as NH 4 Cl. • If an acid is added, NH 3 reacts with the H+. • If a base is added, the NH 4+ ion from the salt reacts with the OH–.

Titration • Titration standard solution – Analytical method in which a standard solution is used to determine the concentration of an unknown solution

Titration • Equivalence point (endpoint) – Point at which equal amounts of H 3 O+ and OH- have been added. – Determined by… • indicator color change • dramatic change in p. H

Titration + O moles H 3 = moles M V = M V M: Molarity V: volume OH

Titration • 42. 5 m. L of 1. 3 M KOH are required to neutralize 50. 0 m. L of HCl. Find the molarity of HCl. H 3 O + M=? V = 50. 0 m. L OHM = 1. 3 M V = 42. 5 m. L MV= MV M(50. 0 m. L) =(1. 3 M)(42. 5 m. L) M = 1. 1 M