There are three ways to define acid base reactions

Bronsted-Lowry definition

Acids will donate protons, bases accept protons

Arrhenius definition

Acids create H+ ions in solutions, bases create OH- ions in solutions

Lewis definition

Acids are electron acceptors, bases are electron donors
Almost everything is an acid base reaction by this definition

Strong acids that completely dissociate in water:

$\ceHCl−>H+andCl$
- Hydro + Chlor + ic acid
$\ceHBR$
- Hydrobromic Acid
$\ceHNO3−>H+NO3$
- Nitric Acid
$\ceHCIO3$
- Chloric Acid
$\ceHClO4$
- Perchloric Acid
$\ceHI$
- Hydroionic Acid
$\ceH2SO4$
- Sulfuric Acid

-ous acids (weak acids) HNO2

NO2=nitrite
Nitrous Acid
$\ce H NO 2 \ce <=> \ce H - + \ce NO 2 +$

Conjugate Acids and Bases

$\ce N H 3 + H 20 <=> H N 4 + O H$

NH3 is a base, H2O is the acid
NH4 is the conjugate acid, OH is the conjugate base

$\ce H NO 2 + H 2 O <=> H 3 O + + NO 2 -$

$\ce H NO 2$ is the acid
$\ce H 2 O$ is the base
$\ce H 3 O +$ is the conjugate acid
$\ce NO 2 -$ is the conjugate base
The stronger acid is $\ce H 3 O +$

\ce H 2 O + <=> H + + O H -

Autoionization of water, Kc = 10

$\ce p [H +] = - l o g ([H +])$ $p H = - lo g (1 0^{- 7}) = 7$ $pO H = - lo g (1 0^{- 7}) = 7$ Sig Figs and pH

Sig figs are based on your concentration of H+ ( $\ce [H +]$ )
ex. 2.37468755 $=>$ 2.375
pH always has a decimal, 7 $\ce - >$ 7.0

$K w$ , $p H$ , and $pO H$

ic→ate ite→ous

All equilibrium constants are just Kc for a specific kind of reaction

pH is defined as $- l o g \ce [H +]$ and $- l o g \ce [O H -]$ , so for water $p H = pO H = 7$

A $p H$ < $7$ means that hydrogen ion concentration has increased and the solution is acidic. A $p H > 7$ means that hydrogen ion concetration has decreased and the solution is basic.

This is only valid for $T = 298 K$

$pO H + p H = 14$ $\ce - l o g ([H +] [O H -]) = - l o g (1 0^{-} 14)$

Concentration combined will always equal 14 in an acid base reaction p of many things can be found, like $p K A$

Is this exothermic or endothermic as written $\ce H 2 O + H <=> O H -$ Exothermic because a bond is broken

Finding pH

Strong Acids

Find the $p H$ of a 1.5M solution of $\ce H ClO 4$

$\ce H ClO 4 - > H + + ClO -$ $\ce [H +] = 1.5 M$ since it is a strong acid $p H = - l o g (1.5) = - 0.18$

Weak Acids

When an acid does not fully dissociate we must treat it as an equilibrium situation $\ce H A <=> H + + A -$ $K_{c} = \ce \frac{[ H + ] [ A - ]}{[ H A ]} = K_{a}$

$\ce H NO 2 + H 2 O <=> NO 2 - + H 3 O +$ $K_{a} = \ce \frac{[ H 3 O + ] [ NO 2 - ]}{H NO 2}$

A 0.10M solution of formic acid has a pH measure of 2.38. Determine $K_{a}$

$\ce H COO H <=> H + + H COO -$ $K_{a} = \ce \frac{[ H + ] [ H COO - ]}{[ H COO H ]}$ $p H = \ce - l o g [H +]$ $\ce [H +] = 1 0^{- 2.38} = 4.2 * 1 0^{- 3} M$ $\ce [H COO H] = 0.10 M$ $K_{a} = \ce \frac{( 4.2 \times 1 0 ^{- 3} ) ( 4.2 \times 1 0 ^{- 3} )}{( 0.10 )} = 1.8 \times 1 0^{- 4}$

$\ce K_{a} - > p H$

The $K_{a}$ of acetic acid is 1.8x10^-5. Determine the pH of a 3% by mass mixture with water of acetic acid. (Density is approximately 1.01g/mL)

Dimensional analysis with concentration

Weak Bases

$\ce B + H 2 O <=> H B + + O H -$

Includes $\ce H 2 O$
$\ce B$ is base, $\ce H 2 O$ is acid
$\ce H B$ is conjugate acid
$\ce O H -$ is conjugate base

$\ce K_{c} = \frac{[ H B + ] [ O H - ]}{[ B ]} = K_{b}$ Relaitonship of $K_{a}$ and $K_{b}$ for Conjugates

$\ce N H 3 + H 2 O <=> N H 4 + + O H -$

B, A, CA, CB

$\ce K_{b} N H 3 = \frac{[ N H 4 + ] [ O H - ]}{[ N H 3 ]}$

$\ce K_{b} N H 4 + = \frac{[ N H 3 + ] [ O H - ]}{[ N H 3 ]}$

Kb * Ka = Kw = [H3O+][OH-]

Polyprotic Acids

$\ce H 2 SO 4 - > H + + H SO 4$

$\ce p Ho f [H 2 SO 4] = 2.00 M$

You can force a polyatomic ion to fully dissociate multiple times if you react it with a strong acid

Acidity of Salt Solutions

When you dissolve an ionic solid into water you create a solution. The properties of the cation and anion will effect the pH level of the solution.

Cations

Cations with a proton will act like weak acids and reduce pH, ex $\ce N H 4 +$

Most metal cations will also cause a decrease in $O H -$ concentration and make the solution acidic. Exceptions to this are group 1 and Ca, Sr, and Ba (OH- is soluble with these metals)

Properties that affect acid strength

There are two main determining factors that determine how strong an acid is

Electronegativity/Polarity of the bond

$\ce H A - > H + + A -$ Where $\ce A -$ is stability

More polar bonds are more acidic

A hydrogen bonded to a strongly electronegative atom will more easily dissociate since it is already fairly positive

Resonance

A polyatomic anion is often very stable because the negative charge can be “shared” by many of the atoms in the anion due to resonance

Common Ion Effect

Consider the case where the same ion is being dissolved into solution from two different sources, such as adding ammonia and ammonium nitrate to a solution

Find the pH of a 0.30M solution of ammonia

$\ce N H 3 + H 2 O <=> N H 4 + + O H -$ $\ce K_{b} = \frac{[ N H 4 + ] [ O H - ]}{[ N H 3 ]}$ $1.8 x 1 0^{- 5} = \frac{x ^{2}}{0.30 - x}$ $x = 0.0023 M = [O H -]$ $pO H = - l o g (.0023) = 2.64$ $p H = 14 - pO H = 11.36$

Find the pH of a 0.30M solution of ammonia and 0.30M solution of ammonium nitrate

$1.8 x 1 0^{- 5} = \frac{( .30 + x ) x}{.30 - x}$ $8 x 1 0^{- 5} = x = [O H -]$ $pO H = - l o g (1.8 x 1 0^{- 5}) = 4.74$ $p H = 14 - pO H = 9.26$

Buffers

A buffer is a solution that can resist changes in pH by containing obth a weak acid and a weak base that neutralizes changes in $\ce H +$ and $\ce O H -$ concentrations from the addition of small amounts of strong acids and bases. It is generally made by mixing weak conjugates of each other, such as $\ce C H 3 COO H$ and $\ce C H 3 COO -$

Henderson Hasselbach Equation

$- l o g (K_{a})$ $p K_{a} = p H - l o g (\ce \frac{[ C H 3 COO - ]}{[ C H 3 COO H ]})$ $p H = p K_{a} + l o g (\ce \frac{[ A - ]}{[ H A ]})$ $pO H = p K_{b} + l o g (\ce \frac{[ H A ]}{[ A - ]})$

Buffer capacity and Range

The pH of a buffer depends on the ratio of the concentrations of the two conjugates but the total amount of acid or base it can neutralize depends on the total concentrations of the two

pH range

In general a buffer is no longer effective when the ratio of the concentrations becomes equal to 10 or 1/10. Since log(10)=1 the general pH range of a buffer is about + or - 1 from the original pH of the buffer

Titrations

A titration is an experimental technique of determining a variety of properties of an acid or a base. They come in three types, strong acid-strong basem weak acid-strong base, strong acid-weak base

Equivalcne point

When you have added equal moles of titrant to the titrand/analyte (equal moles, not equal volumes)

Indicators

Indicators are chemicals that change colors as the pH levels change. You want to pick an indicator that changes color at a pH near the equivalence point.

Doesnt affect the reaction

Solubility in Water

Water is a “universal” solvent

As temperature goes up you can dissolve more of a solid solute in water

solute < maximum = unsaturated solute = maxiumum = saturated solute > maximum = supersaturated

Supersaturation

We can heat water up and dissolve something to saturation in it. The solution is then slowly cooled down, all of the solute will remain dissolved even though it is above the saturation point. This is supersaturation

Super saturated solutions are very delicate and a tiny disturbance will cause the excess solute to precipitate out.

s = solubility of CASO4 s represents solubility

Ice table with 0,0 +s, +s,

If $Ks p = s^{2}$ then you are in a saturated solution

Ks p = [B A 2 +] [F]^{2}

Ks p = s (2 s)^{2} = 4 s^{3}

$K_{s p} = s^{2}$

NItrates are all fully soluble

Ksp = 0.50M +s(2s)^2 4.0x10^-9=4s^3+2s

Acidity of the solution and affect on solubility

A lot of relatively insoluble salts become soluble in solutions that have other solutes that affect the pH of the solution

🪴 Jacob's Notes

Explorer

Acids and Bases

Is this exothermic or endothermic as written $\ce H 2 O + H <=> O H -$ Exothermic because a bond is broken

Finding pH

Strong Acids

Weak Acids

$\ce H NO 2 + H 2 O <=> NO 2 - + H 3 O +$ $K_{a} = \ce \frac{[ H 3 O + ] [ NO 2 - ]}{H NO 2}$

$\ce H COO H <=> H + + H COO -$ $K_{a} = \ce \frac{[ H + ] [ H COO - ]}{[ H COO H ]}$ $p H = \ce - l o g [H +]$ $\ce [H +] = 1 0^{- 2.38} = 4.2 * 1 0^{- 3} M$ $\ce [H COO H] = 0.10 M$ $K_{a} = \ce \frac{( 4.2 \times 1 0 ^{- 3} ) ( 4.2 \times 1 0 ^{- 3} )}{( 0.10 )} = 1.8 \times 1 0^{- 4}$

$\ce K_{a} - > p H$

Weak Bases

Polyprotic Acids

Acidity of Salt Solutions

Cations

Properties that affect acid strength

Common Ion Effect

$\ce N H 3 + H 2 O <=> N H 4 + + O H -$ $\ce K_{b} = \frac{[ N H 4 + ] [ O H - ]}{[ N H 3 ]}$ $1.8 x 1 0^{- 5} = \frac{x ^{2}}{0.30 - x}$ $x = 0.0023 M = [O H -]$ $pO H = - l o g (.0023) = 2.64$ $p H = 14 - pO H = 11.36$

Buffers

Henderson Hasselbach Equation

Buffer capacity and Range

pH range

Titrations

Equivalcne point

Indicators

Solubility in Water

Supersaturation

Explorer

Backlinks

Graph View

Table of Contents

🪴 Jacob's Notes

Explorer

Acids and Bases

Is this exothermic or endothermic as written \ceH2O+H<=>OH− Exothermic because a bond is broken

Finding pH

Strong Acids

Weak Acids

\ceHNO2+H2O<=>NO2−+H3O+ Ka​=\ceHNO2[H3O+][NO2−]​

\ceHCOOH<=>H++HCOO− Ka​=\ce[HCOOH][H+][HCOO−]​ pH=\ce−log[H+] \ce[H+]=10−2.38=4.2∗10−3M \ce[HCOOH]=0.10M Ka​=\ce(0.10)(4.2×10−3)(4.2×10−3)​=1.8×10−4

\ceKa​−>pH

Weak Bases

Polyprotic Acids

Acidity of Salt Solutions

Cations

Properties that affect acid strength

Common Ion Effect

\ceNH3+H2O<=>NH4++OH− \ceKb​=[NH3][NH4+][OH−]​ 1.8x10−5=0.30−xx2​ x=0.0023M=[OH−] pOH=−log(.0023)=2.64 pH=14−pOH=11.36

Buffers

Henderson Hasselbach Equation

Buffer capacity and Range

pH range

Titrations

Equivalcne point

Indicators

Solubility in Water

Supersaturation

Explorer

Backlinks

Graph View

Table of Contents

Is this exothermic or endothermic as written $\ce H 2 O + H <=> O H -$ Exothermic because a bond is broken

$\ce H NO 2 + H 2 O <=> NO 2 - + H 3 O +$ $K_{a} = \ce \frac{[ H 3 O + ] [ NO 2 - ]}{H NO 2}$

$\ce H COO H <=> H + + H COO -$ $K_{a} = \ce \frac{[ H + ] [ H COO - ]}{[ H COO H ]}$ $p H = \ce - l o g [H +]$ $\ce [H +] = 1 0^{- 2.38} = 4.2 * 1 0^{- 3} M$ $\ce [H COO H] = 0.10 M$ $K_{a} = \ce \frac{( 4.2 \times 1 0 ^{- 3} ) ( 4.2 \times 1 0 ^{- 3} )}{( 0.10 )} = 1.8 \times 1 0^{- 4}$

$\ce K_{a} - > p H$

$\ce N H 3 + H 2 O <=> N H 4 + + O H -$ $\ce K_{b} = \frac{[ N H 4 + ] [ O H - ]}{[ N H 3 ]}$ $1.8 x 1 0^{- 5} = \frac{x ^{2}}{0.30 - x}$ $x = 0.0023 M = [O H -]$ $pO H = - l o g (.0023) = 2.64$ $p H = 14 - pO H = 11.36$