Ch. 13 Acids and Bases

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Acid-Base Theory

Arrhenius Theory
- acids as substances that increase the concentration of $\ce{H+}$ or $\ce{H3O+}$ ions in aqueous solution
- bases as substances that increase the concentration of $\ce{OH-}$ ions in aqueous solution
Brønsted-Lowry Theory
- acids as proton donors
- bases as proton acceptors
- solves the problem of substances without $\ce{OH}$ , such as $\ce{NH3}$ , increasing concentration of $\ce{OH-}$
modern theory
- acids have one or more hydrogen atoms weakly bound to the rest of the molecule
  - hydrogen ions ionize in aqueous solution
  - can begin with $\ce{H}$ or end with $\ce{-COOH}$ functional group for organic acids

Nomenclature

binary acids contain hydrogen and one other atom
- named hydro[atom root]ic acid, e.g. $\ce{HF}$ hydrofluoric acid
polyatomic anions in the acid have changed endings
- -ate becomes -ic and -ite bceomes -ous
organic acids end in -oic
hydroxide bases contain hydroxide ions
- named with the name of the metal with roman numerals, then hydroxide
nitrogen bases contain nitrogen; are related to ammonia and are called amines
- hydrogen in ammonia replaced with methyl groups ( $\ce{-CH3}$ ), ethyl groups ( $\ce{-CH2CH3}$ ), etc.

Acids

strong acid: acid that completely ionizes in aqueous
- $\ce{HCl}$ , $\ce{HBr}$ , $\ce{HI}$ , $\ce{HClO4}$ , $\ce{HNO3}$ , $\ce{H2SO4}$ (only first hydrogen in sulfuric acid)
weak acid: all other acids; partially ionizes
acid strength and bond strength are inversely proportional
- increases going left to right and top to bottom
oxoacids: mineral acids that are not binary; contain hydrogen, oxygen, and another element
- oxygen bonded to central atom, hydrogen bonded to oxygen
- for same central atom and hydrogens, strength and stability increases as number of oxygen increases
- for same oxygen and hydrogen, strength increases as electronegativity increases
- the weaker the $\ce{OH}$ bond, the stronger the acid

Bases

strong base: all metal hydroxides, though only some are soluble (group IA)
weak base: bases related to ammonia
- base strength decreases as electronegativity of organic functional groups increases (reverse of acids)

Anhydrides

compounds that become common acids and bases when added to water
- acid anhydriddes: oxides of nonmetals
  - $\ce{SO3 + H2O -> H2SO4}$
- basic anhydrides: oxides of metals
  - $\ce{K2O + H2O -> 2KOH}$

Neutralization Reactions

reaction between acids and bases
forms a salt and water
$\ce{HBR(aq) + KOH(aq) -> KBr(aq) + H2O(l)}$
(acid) (base) (salt) (water)
acids/bases can react with acid/base anhydrides

Polyprotic Acids

acids containing more than one ionizing hydrogen
- e.g. $\ce{H2SO4}$ , $\ce{H3PO4}$
all but sulfuric acid (first ionizes completely, second partially) are weak acids
product formed depend on amount of base used
- $\ce{H3PO4 + OH- -> H2PO4^- + H2O}$
- $\ce{H3PO4 + 2OH- -> HPO4^2- + 2H2O}$
- $\ce{H3PO4 + 3OH- -> PO4^3- + 3H2O}$

Brønsted-Lowry Theory

conjugate acid-base pairs: compounds that differ in formulas by one $\ce{H+}$ $H X^{+}$
- $\ce{A1 + B2 -> B1 + A2}$ , $\ce{A1}$ and $\ce{B1}$ are a conjugate pair, $\ce{A2}$ and $\ce{B2}$ are a conjugate pair
- $\ce{Conjugate acid <=> Conjugate base + H+}$ $C o n j u g a t e a c i d C o n j u g a t e b a s e + H X^{+}$ , stronger one reacts more (less of it at equilibrium)
  - same logic can be applied to a complete acid/base chemical reaction
- $\ce{H3O+}$ $H X_{3} O X^{+}$ is the strongest acid and $\ce{OH-}$ $O H X^{-}$ is the strongest base in aqueous solution
  - acids/bases stronger than $\ce{H2O}$ react with water to produce $\ce{H3O+}$ / $\ce{OH-}$

pH and pOH

pH: measure of molar hydrogen ion concentration
- $\ce{pH}=-\log{[\ce{H+}]}=-\log{[\ce{H3O+}]}$
- goes from 0 to 14, with lower numbers being more acidic
pOH: measure of molar hydroxide ion concentration
- $\ce{pOH}=-\log{[\ce{OH-}]}$
equilibrium expression for equation $\ce{H2O <=> H+ + OH-}$ $H X_{2} O H X^{+} + O H X^{-}$ is $K_w=[\ce{H+}][\ce{OH-}]=1.0\cdot10^{-14}$ , where $K_w$ $K_{w}$ is the autoionization constant of water
- $\ce{pH}$ and $\ce{pOH}$ relate by $\ce{p}K_w=\ce{pH}+\ce{pOH}=14.00$
$\ce{pH}<7$ $p H < 7$ is acidic, $\ce{pH}>7$ $p H > 7$ is basic, $\ce{pH}=7$ $p H = 7$ is neutral at standard temperature $(25^\circ\pu{C})$ $(2 5^{\circ} C)$
- higher temperature lowers neutral $\pu{pH}$
for strong acids, $[\ce{H+}]=[\text{strong acid}]$ ; for strong bases, $[\ce{OH-}]=[\text{strong base}]$

Weak Acid/Base pH

in equation $\ce{HF + H2O <=> H3O+ + F-}$ or equivalently $\ce{HF <=> H+ + F-}$ :
$K_a=\frac{[\ce{H3O+}][\ce{F-}]}{[\ce{HF}]}=\frac{[\ce{H+}][\ce{F-}]}{[\ce{HF}]}$
and $\ce{pH}=-\log{[\ce{H+}]}=-\log{[\ce{H3O+}]}$
$K_a$ is acid dissociation constant: the equilibrium constant for the acid dissosciation equation

Example

A solution contains $\pu{0.100 } M$ $\pu{\ce{HF}}$ . What is the pH of the solution given $K_a=6.6\cdot10^{-4}$ ?

Use an RICE table

Reaction	$\ce{HF}$	$\ce{H+}$	$\ce{F-}$
Initial	$\pu{0.100 }M$	$\pu{0 }M$	$\pu{0 }M$
Change	$-x$	$+x$	$+x$
Equilibrium	$0.100-x$	$x$	$x$

The equilibrium expression gives
$K_a=\frac{[\ce{H+}][\ce{F-}]}{[\ce{HF}]}\rightarrow 6.6\cdot10^{-4}=\frac{x^2}{0.100-x}$
$K_a$ is small enough that the $x$ in the denominator can be ignored
$6.6\cdot10^{-4}=\frac{x^2}{0.100}\rightarrow x=8.12\cdot10^{-3}$
Finally, since $[\ce{H+}]=x=8.12\cdot10^{-3}$ , the $\ce{pH}$ is $-\log{8.12\cdot10^{-3}}=2.09$

$K_b$ is base dissociation constant: the equilibrium constant for base dissociation

Example

What is the pH of a $\pu{0.500 } M$ solution of $\ce{CH3NH2}$ given $K_b=4.2\cdot10^{-4}$ ?

This is an ammonium-related base, so the reaction contains water as the acid
Using a RICE table

Reaction	$\ce{CH3NH2}$	$\ce{H2O}$	$\ce{CH2NH3+}$	$\ce{OH-}$
Initial	$\pu{0.500 }M$	-	$\pu{0 }M$	$\pu{0 M}$
Change	$-x$	-	$+x$	$+x$
Equilibrium	$0.500-x$	-	$x$	$x$

The equilibrium expression gives
$K_b=\frac{[\ce{OH-}][\ce{CH2NH3+}]}{[\ce{CH3NH2}]}\rightarrow4.2\cdot10^{-4}=\frac{x^2}{0.500-x}$
$K_b$ is small enough that the $x$ in the denominator can be ignored
$4.2\cdot10^{-4}=\frac{x^2}{0.500}\rightarrow x=1.45\cdot10^{-2}$
Since $[\ce{OH-}]=x=1.45\cdot10^{-2}$ , the $\ce{pOH}$ is $-\log{1.45\cdot10^{-2}}=1.84$
Finally, since $\ce{pH + pOH}=14$ , the $\ce{pH}$ is $14-1.84=12.16$

$K_aK_b=K_w=1.00\cdot10^{-14}$

Hydrolysis Reaction

if the anion of a salt is the conjugate base of a weak acid, it will react with water, forming the acid in a basic solution
- $\ce{F- + H2O <=> HF + OH-}$
if the cation of a salt is the conjugate acid of a weak base, it will react with water, forming the base in an acidic solution
- $\ce{NH4^+ + H2O <=> NH3 + H3O^+}$
from the above, a salt in an aqueous solution produces an acidic, basic, or neutral solution
- neutral only if strong acid/strong base
- solve for pH using the ion that is conjugate to a weak acid/base and using the same process

Example

When $\ce{Na2SO3}$ is dissolved in water, it ionizes into $\ce{Na+}$ cations and $\ce{SO3^2-}$ anions.
These are conjugate pairs to $\ce{NaOH}$ (strong base) and $\ce{H2SO3}$ (weak acid) respectively.
A strong base is stronger than a weak acid, so the solution will be basic

Buffer Solutions

solutions that contain a conjugate acid-base pair to prevent pH from changing too much from addition of strong acid/base
solution contains some conjugate acid/base so less $[\ce{H+}]$ or $[\ce{OH-}]$ is produced

Example

Find the pH of a solution of $\pu{0.125 }M$ $\ce{N2H4}$ and $\pu{0.321 }M$ $\ce{N2H5Cl}$ given $K_b=9.6\cdot10^{-7}$

Using a RICE table

Reaction	$\ce{N2H4}$	$\ce{H2O}$	$\ce{N2H5+}$	$\ce{OH-}$
Initial	$\pu{0.125 }M$	-	$\pu{0.321 }M$	$\pu{0 }M$
Change	$-x$	-	$+x$	$+x$
Equilibrium	$0.125-x$	-	$0.321+x$	$x$

The equilibrium expression gives
$K_b=\frac{[\ce{OH-}][\ce{N2H5+}]}{[\ce{N2H4}]}\rightarrow 9.6\cdot10^{-7}=\frac{x(0.321+x)}{0.125-x}$
$K_b$ is small enough that it may be ignored.
$9.6\cdot10^{-7}=\frac{0.321x}{0.125}\rightarrow x=3.74\cdot10^{-7}$
Since $[\ce{OH-}]=x=3.74\cdot10^{-7}$ , $\ce{pOH}=-\log{3.74\cdot10^{-7}}=6.43$ and $\ce{pH}=14-6.43=7.57$

Polyprotic Acids

each dissociation constant $K_{an}$ is smaller than the previous $K_{a(n-1)}$
only the first dissociation step is necessary to determine pH

Henderson-Hasselbalch Equation

derived from taking the negative log of both sides of the equilibrium expressions
$\ce{pH}=\ce{p}K_a+\log{\frac{[\text{conjugate base}]}{[\text{conjugate acid}]}}$
$\ce{pOH}=\ce{p}K_b+\log{\frac{[\ce{conjugate acid}]}{[\text{conjugate base}]}}$

Titration Curves

graph of pH vs volume of titrant added

start with an acid or base
titrant is added until the equivalence point: all reactant is converted into product; salt solution
more titrant is added, leaving excess titrant

middle part where pH drastically changes is equivalence point
polyprotic acids produce multiple equivalence points
at half equivalence point $\ce{pH}=\ce{p}K_a$ or $\ce{pOH}=\ce{p}K_b$
flatter parts are buffer solutions, since they contain a conjugate acid-base mixture