Chapter+13,+Section+2

=  =  S11.A.1.3.1  Use appropriate quantitative data to describe or interpret change in systems (e.g., biological indices, electrical circuit data, automobile diagnostic systems data). =Acid Base Character= =For a molecule with a H-X bond to be an acid, the hydrogen must have a positive [|oxidation number] so it can ionize to form a positive +1 ion. For instance, in sodium hydride (NaH) the hydrogen has a -1 charge so it is not an acid but it is actually a base. Molecules like CH4 with nonpolar bonds also cannot be acids because the H does not ionize. Molecules with strong bonds (large [|electronegativity] differences), are less likely to be strong acids because they do not ionize very well. For a molecule with an X-O-H bond (also called an [|oxoacid]) to be an acid, the hydrogen must again ionize to form H+. To be a base, the O-H must break off to form the hydroxide ion (OH-). Both of these happen when dealing with oxoacids.= = = HCl, HBr, HI, H2SO4, HNO3, HClO3, HClO4 To calculate a pH value, it is easiest to follow the standard "Start, Change, Equilibrium" process. Example Problem: Determine the pH of a 0.25 M solution of HBr. HA(aq) H+(aq) + A-(aq) The equilibrium constant for the dissociation of an acid is known as **K**a. The larger the value of **K**a, the stronger the acid.
 * Strong Acids:** These acids completely ionize in solution so they are always represented in chemical equations in their ionized form. There are only seven (7) strong acids:
 * **Answer:** ||
 * || HBr (aq) || → || H+(aq) || + || Br-(aq) ||
 * Start: || .25 M ||  || 0 M ||   || 0 M ||
 * Change: || -.25 ||  || +.25 ||   || +.25 ||
 * Equilibrium: || 0 ||  || .25 ||   || .25 ||
 * |||||||| pH = -log[H+] = -log(.25) = 0.60 ||
 * Weak Acids:** These are the most common type of acids. They follow the equation:
 * Ka = || [H+][A-]

[HA] || Example Problem: Determine the pH of 0.30 M acetic acid (HC2H3O2) with the **K**a of 1.8x10-5.
 * **Answer:** ||
 * Write an equilibrium equation for the acid: ||
 * HC2H3O2 [[image:http://www.shodor.org/unchem/basic/ab/SYMB/DAR.GIF width="20" height="10" align="center"]]

H+ + C2H3O2- ||
 * Write the equilibrium expression and the Ka value: ||
 * || Ka = || [H+][C2H3O2-]

[HC2H3O2] || = 1.8x10-5 ||  || and solve for [H+]: ||
 * "Start, Change, Equilibrium": ||
 * ||  || HC2H3O2 || [[image:http://www.shodor.org/unchem/basic/ab/SYMB/DAR.GIF width="20" height="10" align="center"]] || H+ || + || C2H3O2- ||
 * Start: || 0.30 M ||  || 0 M ||   || 0 M ||   ||
 * Change: || -x ||  || +x ||   || +x ||
 * Equilibrium: || 0.30 - x ||  || x ||   || x ||   ||
 * Substitute the variables (disregard the "-x" because it is so small compared to the 0.30)
 * || Ka = 1.8x10-5 = || (x)(x)

(.30 - x) || = || x2

.30 ||  || LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2 Example Problem: Determine the pH of a 0.010 M solution of Ba(OH)2. Weak Base + H2O conjugate acid + OH- example: NH3 + H2O NH4+ + OH-
 * x = [H+] = 2.3x10-3 ||
 * pH = -log[H+] = 2.64 ||
 * Strong Bases:** Like strong acids, these bases completely ionize in solution and are always represented in their ionized form in chemical equations. There are only eight (8) strong bases:
 * **Answer:** ||
 * || Ba(OH)2(aq) || → || Ba2+(aq) || + || 2OH-(aq) ||
 * Start: || .010 M ||  || 0 M ||   || 0 M ||
 * Change: || -.010 ||  || +.010 ||   || +.010 ||
 * Equilibrium: || 0 ||  || .010 ||   || .010 ||
 * pOH = -log[OH-] = -log(.010) = 1.70 ||
 * pH = 14.00 - 1.70 = 12.30 ||
 * Weak Bases:** These follow the equation:


 * K**b is the base-dissociation constant:
 * Kb = || [conjugate acid][OH-]

[weak base][H2O] ||
 * example: Kb = || [NH4+][OH-]

[NH3[H2O] || To calculate the pH of a weak base, we must follow a very similar "Start, Change, Equilibrium" process as we did with the weak acid, however we must add a few steps. Example Problem: Determine the pH of 0.15 M ammonia (NH3) with a **K**b=1.8x10-5.
 * K**a x **K**b = **K**w = 1.00x10-14
 * **Answer:** ||
 * Write the equilibrium equation for the base: ||
 * NH3 + H2O [[image:http://www.shodor.org/unchem/basic/ab/SYMB/DAR.GIF width="20" height="10" align="center"]]

NH4+ + OH- ||
 * Write the equilibrium expression and the Kb value: ||
 * || Kb = || [NH4+][OH2-]

[NH3][H2O] || = 1.8x10-5 ||  || and solve for [OH-]: ||
 * "Start, Change, Equilibrium": ||
 * ||  || NH3 || + || H2O || [[image:http://www.shodor.org/unchem/basic/ab/SYMB/DAR.GIF width="20" height="10" align="center"]] || NH4+ || + || OH- ||
 * Start: || 0.15 M ||  || -- ||   || 0 M ||   || 0 M ||   ||
 * Change: || -x ||  || -- ||   || +x ||   || +x ||
 * Equilibrium: || 0.15 - x ||  || -- ||   || x ||   || x ||   ||
 * Substitute the variables (disregard the "-x" because it is so small compared to the 0.15)
 * || Kb = 1.8x10-5 = || (x)(x)

(.15 - x) || = || x2

.15 ||  || When dealing with weak acids and weak bases, you also might have to deal with the "common ion effect". This is when you add a salt to a weak acid or base that contains one of the ions present in the acid or base. To be able to use the same process to solve for pH when this occurs, all you need to change are your "start" numbers. Add the [|molarity] of the ion, which comes from the salt, and then solve the **K**a or **K**b equation as you did earlier. Example Problem: Find the pH of a solution formed by dissolving 0.100 mol of HC2H3O2 with a **K**a of 1.8x10-8 and 0.200 mol of NaC2H3O2 in a total volume of 1.00 L.
 * x = [OH-] = 1.6x10-3 M ||
 * pOH = -log[OH-] = 2.80 ||
 * pH = 14.00 - 2.80 - 11.20 ||
 * **Answer:** ||
 * || HC2H3O2(aq) || → || H+(aq) || + || C2H3O2-(aq) ||
 * Start: || .10 M ||  || 0 M ||   || .20 M ||
 * Change: || -x ||  || +x ||   || +x ||
 * Equilibrium: || .10 - x ||  || x ||   || .20 + x ||
 * || Ka = 1.8x10-8 = || (x)(.20 + x)

(.10 - x) || = || (x)(.20)

(.10) ||  ||
 * **x = [H+] = -9.0x10-9** ||
 * **pH = -log(9.0x10-9) = 8.05** ||

Pure water is both a weak acid and a weak base. By itself, water forms only a very small number of the H3O+ and OH- ions that characterize aqueous solutions of stronger acids and bases. ||  ||  H3O+(//aq//)  ||   ||  +  ||   ||  OH-(//aq//)  ||   ||   ||
 * H2O(//l//) ||   ||  +  ||   ||  H2O(//l//)  ||   ||  [[image:http://www.cartage.org.lb/en/themes/sciences/chemistry/inorganicchemistry/acidsbases/measure/equilibr.gif width="17" height="9" align="center"]]


 * base ||   ||   ||   ||  acid  ||   ||   ||   ||  acid  ||   ||   ||   ||  base  ||   ||   ||

The concentrations of the H3O+ and OH- ions in water can be determined by carefully measuring the ability of water to conduct an electric current. At 25oC, the concentrations of these ions in pure water is 1.0 x 10-7 moles per liter. [H3O+] = [OH-] = 1.0 x 10-7 //M// (at 25°C) When we add a strong acid to water, the concentration of the H3O+ ion increases. HCl(//aq//) + H2O(//l//) H3O+(//aq//) + Cl-(//aq//) At the same time, the OH- ion concentration decreases because the H3O+ ions produced in this reaction neutralize some of the OH- ions in water. H3O+(//aq//) + OH-(//aq//) 2 H2O(//l//) The product of the concentrations of the H3O+ and OH- ions is constant, no matter how much acid or base is added to water. In pure water at 25oC, the product of the concentration of these ions is 1.0 x 10-14.

[H3O+][OH-] = 1.0 x 10-14

The range of concentrations of the H3O+ and OH- ions in aqueous solution is so large that it is difficult to work with. In 1909 the Danish biochemist S. P. L. Sorenson suggested reporting the concentration of the H3O+ ion on a logarithmic scale, which he named the **pH scale**. Because the H3O+ ion concentration in water is almost always smaller than 1, the log of these concentrations is a negative number. To avoid having to constantly work with negative numbers, Sorenson defined pH as the negative of the log of the H3O+ ion concentration. pH = -log [H3O+] The concept of pH compresses the range of H3O+ ion concentrations into a scale that is much easier to handle. As the H3O+ ion concentration decreases from roughly 100 to 10-14, the pH of the solution increases from 0 to 14. If the concentration of the H3O+ ion in pure water at 25oC is 1.0 x 10-7 //M//, the pH of pure water is 7. pH = -log [H3O+] = -log (1.0 x 10-7) = 7 When the pH of a solution is less than 7, the solution is acidic. When the pH is more than 7, the solution is basic.
 * Acidic: ||   ||   ||   ||  pH < 7  ||


 * Basic: ||   ||   ||   ||  pH > 7  ||