 • Introduction
• pH and pOH
• Kw, Ka, and Kb
• Strong acids and bases
• Percent ionization example
• Titrations
• Titration graphs
• Buffers/Henderson Hasselbalch

#### Introduction

The study of acids and bases involves determining concentrations of H+ and OH- ions within solutions. There are three different ways to define acids and bases. The Arrhenius definition states that acids are substances that dissociate (separate into ions) in water to give H+ ions. Bases are substances that dissociate in water to form hydroxide (OH–) ions. The Bronsted-Lowry definition states that acids are proton (H+) donors and bases are proton acceptors. The Lewis definition states that acids are electron-pair acceptors and bases are electron pair donors.

## The Basics #### pH and pOH

pH is a way to measure the concentration of H+ ions in a solution. pH = -log[H+]. pH is measured on a scale from 0-14. A lower pH means that a solution has a higher concentration of H+ ions and is therefore more acidic. A higher pH means that the concentration of H+ is smaller, meaning that a solution is more basic. Like pH, pOH measures the concentration of OH- in solution. pOH = -log[OH-] and pH = 14 - pOH. Therefore, if the concentration of either H+ or OH- are known, the other concentration is known as well.

#### Kw

Kw is a form of the equilibrium constant (more on the equilibrium constant on the equilibrium page). Given the reaction of the self ionization of water, H2O (l) → H+ + OH-, Kw equals [H+][OH-]. At 25 degrees Celsius, Kw = 1.0 x 10^14. Since pH = pOH in pure water, this means that the both the pH and pOH of water at 25 degrees celsius is 7. Because Kw is temperature dependent, the pH of water will slightly change at different temperatures.

#### Strong acids and bases

It is important to memorize the list of strong acids and bases. Acids and bases are strong if they dissociate or ionize completely in water. This means that if the solid acid or base is added to water, all of the ions will separate (as long as the solution is still unsaturated). With weak acids and bases, even in an unsaturated solution, not all the ions will separate. The strong acids are: HCl, HBr, HI, HNO3, H2SO4, HClO3, HClO4. The strong bases are LiOH, NaOH, KOH, RbOH, CsOH,Ca(OH)2, Sr(OH)2, Ba(OH)2. The strong bases are the group 1 hydroxides along with the Ca, Sr, and Ba hydroxides which makes them easier to remember.

## Acid Base Equilibrium

#### pKa and pKb Since weak acids don’t ionize completely in water, they create an equilibrium between their un-ionized acid and its conjugate base. This equilibrium constant has the same calculation of concentration of products divided by concentration of reactants (all raised to stoichiometric powers) as the rest of the equilibrium constants. Given that acids ionize based on the general equation HA → H3O+ + A-, the equation for Ka is pictured above. Note that H3O+ and H+ can be interchanged for notational purposes.

#### pKa and pKb

pKa and pKb are calculated by taking the negative logarithm (-log) of either Ka or Kb. These numbers have many applications which are explored further below.

#### Weak Bases Bases form an equilibrium just like weak acids. Rather than using Ka, Kb is used for bases. Given that bases ionize based on the general equation B → OH- + HB+, the equation for Kb is pictured above. ## Percent Ionization and pH Example

Often in calculations with Ka and Kb, small numbers can be ignored as they greatly complicate the math and cause negligible change in the final answer. As a rule of thumb, a number that is 1000 times smaller than a number it is added to can be crossed out. The example above shows one such situation where this rule is used, this case to find percent ionization of a weak acid.

## Titrations #### Introduction

Titrations are a laboratory technique used to determine pH using stoichiometry, measurement, and other mathematical techniques to determine H+ or OH- concentrations. There are four possible titration combinations: strong acid and strong base, strong acid and weak base, weak acid and strong base, and weak acid and weak base.

#### Steps of a titration

The best way to learn about titrations is to actually do it! But if that is impossible, here is a general overview of how a titration works. The materials needed are a burette (long glass tube with stopcock on the end), stand to hold up the burette, beaker or flask, the acid and base solutions, and an appropriate indicator. Add a known volume of acid or base to the beaker and add a few drops of the proper indicator. Then choose an appropriate chemical to place in the burette (a strong acid for bases in the beaker or a strong base for acids in the beaker). Measure the starting volume in the burette, making sure to count down from the marked line above the measurement (the markings are opposite that of a graduated cylinder). Then slowly add the titrant to the beaker while stirring. As soon as the indicator changes color, stop the titration and note the final volume in the burette.

It is important to note that the inside of the beaker or flask can be slightly wet as the addition of water will not change the total number of moles of solute within the solution. However, the inside of the burette must contain only the titrant and no other liquids because the presence of water will cause the measurement of the volume of titrant needed to be larger than it should be as the concentration of the titrant is slightly reduced.

#### Strong Acid and Strong Base

The net ionic equation for any strong acid and strong base is H+ + OH- → H2O. Therefore, the moles of either H+ or OH- used in the titrant (depending on whether it is an acid or base) will be equal to the number of moles of the other ion. Using stoichiometry and calculations for concentration, the initial concentration or pH can be found. The final pH of the titration will be 7. Remember that molarity = moles / liters which can be used to determine the moles present in a certain volume of solution of known concentration.

#### Strong Acid and Weak Base

A strong acid and weak base react with the general net ionic equation B + H+ → HB+ + H2O. Usually in a titration the strong acid will be in excess, so the final pH is the -log of the concentration of excess H+ ions. This value will always be below 7 (acidic). If the weak base is in excess, a buffer solution will be formed and the Henderson Hasselbalch equation can be used (more below).

#### Weak Acid and Strong Base

A weak acid and strong base react with the general net ionic equation HA + OH-  →  A- + H2O. Usually in a titration the strong base will be in excess, so final pH can be found by determining the moles of excess OH- and finding pOH, then subtracting that value from 14 to yield pH. This value will always be above 7 (basic). If the weak acid is in excess, a buffer solution will be formed and the Henderson Hasselbalch equation can be used (more below).

#### Weak Acid and Weak Base

Performing a titration with a weak acid and weak base is just silly since the math becomes overly complicated. However, it is important to note that if Ka > Kb, the solution will be acidic, and if Kb > Ka, the solution will be basic.

## Titration Graphs and Calculations #### Background

Titration graphs generally take on the same shape, so plotting a few points is usually enough to sketch the curve. The vertical axis of a titration graph shows the pH of the solution, while the horizontal axis shows the volume of the titrant added. The calculations are slightly different depending on the strengths of acids or bases used.

#### Strong acid and strong base (acid in the beaker, base as titrant)

• Initially, the pH will be acidic. Use -log[H+] to find the pH.
• Before the equivalence point (including the midpoint), the solution is in a buffer phase. This becomes a limiting reactant problem to find how many moles are in excess. Before the equivalence point, the acid will always be in excess because not enough base has been added to neutralize it yet. After finding moles of excess acid and dividing by volume to find concentration, use -log[H+] to find pH.
• At the equivalence point, the pH is 7 since the moles of H+ and OH- are equal.
• After the equivalence point, start by calculating how many moles of base are in excess using moles added - moles used. Then divide by the total volume, making sure to add both the initial volume and the volume of titrant added. Then use -log[OH-] to find pOH and subtract this value from 14 to find pH.
• Titrations of strong acid and strong base with the acid as the titrant can be performed in the same way, switching calculations using base added to acid added. Use 14-log[OH-] to convert from pOH to pH.

#### Weak acid and strong base (acid in the beaker, base as titrant)

• To find initial pH, start by looking up the Ka of the weak acid. Given the general equation HA → H+ + A-, Ka is given by [H+][A-]/[HA]. Since H+ and A- increase by the same amount (which can be seen in a RICE box), the calculation Ka = [x]^2/[acid] can be used. Since x is the concentration of H+, use -log(x) to find pH after solving for x.
• At the midpoint, pH = pKa of the acid in the beaker because [HA] = [A-].
• At the equivalence point, the pH will be above 7. First, calculate the moles of OH- added using molarity*liters. At the equivalence point, this is equal to the moles of H+ used. Now, since the solution is basic, we need to follow the equilibrium equation A- + H2O → OH- + HA. This means that Kb = [OH-][HA]/[A-]. Since HA and OH- increase by the same amount (which can be seen in a RICE box), the calculation Kb = [x]^2/[A-] can be used. Since x is the concentration of OH-, use -log(x) to find pOH after solving for x, then subtract this value from 14 to find pH.
• After the equivalence point, start by calculating how many moles of base are in excess using moles added - moles used. Then divide by the total volume, making sure to add both the initial volume and the volume of titrant added. Then use -log[OH-] to find pOH and subtract this value from 14 to find pH.

#### Strong acid and weak base (base in the beaker, acid as titrant)

• To find initial pH, start by looking up the Kb of the weak base. Given the general equation WB + H2O → OH- + WA, Kb is given by [OH-][WA]/[WB]. Since WA and OH- increase by the same amount (which can be seen in a RICE box), the calculation Kb = [x]^2/[WB] can be used. Since x is the concentration of OH-, use -log(x) to find pOH after solving for x, then subtract this value from 14 to find pH.
• At the midpoint, pOH = pKb of the acid in the beaker because [WB] = [WA+]. Subtract pKb from 14 to find pH.
• At the equivalence point,the pH will be below 7. First, calculate the moles of H+ added using molarity*liters. At the equivalence point, this is equal to the moles of OH- used. Now, since the solution is acid, we need to follow the equilibrium equation WA + H2O → H+ + WB. This means that Ka = [H+][WB]/[WA]. Since H+ and WB increase by the same amount (which can be seen in a RICE box), the calculation Ka = [x]^2/[WA] can be used. Since x is the concentration of H+, use -log(x) to find pH after solving for x.
• After the equivalence point, start by calculating how many moles of acid are in excess using moles added - moles used. Then divide by the total volume, making sure to add both the initial volume and the volume of titrant added. Then use -log[H+] to find pH.

## Buffers

#### Henderson Hasselbalch

Buffers are solutions made from an acid and its conjugate base or a base and its conjugate acid that are difficult to change the pH of. When an acid or base is added to the buffer, the reaction will shift right or left, thus “consuming” the acid or base and keeping the initial pH. A buffer solution can be identified when the ratio between the conjugate acid and conjugate base is between .1 and 10. In other words, the acid and base must be present in a concentration of at most 10 times the other solute. To find the pH a buffer solution can most effectively be used for, add +/- 1 to the pKa of the acid in the buffer.

#### Henderson Hasselbalch

pH = pKa + log([base]/[acid]) The Henderson Hasselbalch equation can be used to calculate changes in pH of a buffer solution. This equation is handy because without it we would have to use Ka or Kb and solve more complicated equilibrium expressions to find pH. Make sure to use concentrations within the log and that the concentrations are within 10 times of one another.