Acids and Bases: Acid/Base Theory Mrs. Crowley’s Chemistry Class

Measuring pH How do we measure pH? Own a pool? Own a spa? How do you check the water quality? Look at my poster. Also…pH meters can do the trick, too:

Indicators Acid-base indicators: This is usually a solution that you add to a solution that will undergo a chemical reaction at a certain pH. The reaction is called “dissociation.” This type of reaction is an equilibrium. Remember, equilibriums go from reactants to products and products to reactants (back and forth). H 2 O H + + OH - This is an example of an “equilibrium” The rate of flow towards products is the same as the rate to reactants.

Check out my poster! There are many different indicators available for chemists. Sometimes, we can combine these indicators to make a broad spectrum of color changes. This is how traditional pH paper is made.

Acid-Base Theories Well, now it’s time to learn what his theory of acids and bases were. Arrhenius’s main contribution to physical chemistry was his theory (1887) that electrolytes, certain substances that dissolve in water to yield a solution that conducts electricity, are separated, or dissociated, into electrically charged particles, or ions, even when there is no current flowing through the solution

Svante Arrhenius Svante said that acids are hydrogen- containing compounds that ionize to yield hydrogen ions (H+) in aqueous solution. He also said that bases are compounds that ionize to yield hydroxide ions (OH-) in aqueous solution.

Acids – further explaination Not all compounds that contain hydrogen are acids! (poor Arrhenius!...he didn’t know this…YET) Also, not all hydrogens in an acid are released as hydrogen ions. Only the hydrogens in very polar bonds are ionizable. EXAMPLE:HCl + H 2 O  H 3 O + + Cl - Highly Polar molecule

Bronsted-Lawry Acids and Bases The next theory on acids and bases came from these two men around the year 1923.

Bronsted-Lowry Theory These guys re-defined an acid as a hydrogen-ion donor. They also re-defined a base as a hydrogen ion acceptor. Let me explain:

Explanation When ammonia (a well known base) combines with water, the ammonia will act as a base. But ammonia doesn’t have an OH! How can it possibly be a base if there is no OH ion? The OH ion will come from the water! Yes…that’s right. The water will donate a hydrogen to the ammonia. Poor Arrehenius thought ammonia’s formula was NH 4 OH! But the molar mass of ammonia was discovered to be MUCH less than that of NH 4 OH!

Further Explanation Does that mean that the water is an acid? Well, according to the Bronsted- Lowry theory…YES! Water acts like a Bronsted-Lowry base. HOW? Look at that “double arrow.”

Equilibriums If the reaction between water and ammonia goes in the reverse direction, NH 4 + will react with OH - to form NH 3 and H 2 O. acid = hydrogen-ion donor. base = hydrogen ion acceptor.

Conjugate Acids and Bases If the reaction goes in the reverse direction, NH 4 + gives up a hydrogen ion. That makes it an Bronsted-Lowry ACID! OH - will accept that hydrogen ion from NH 4 +. That makes it a Bronsted-Lowry BASE! Since NH 4 + is on the right side of the equation, it is called a conjugate acid. Since OH- is on the right side of the equation, it is called a conjugate base.

Definitions Conjugate acids are the particles formed when a base gains a hydrogen ion. A conjugate base is the particle that remains when an acid has donated a hydrogen ion.

Lewis Acids and Bases The last theory for acids and bases came from Gilbert Lewis ( ).

Lewis Acids-Bases A lewis acid is a substance that can accept a pair of electrons to form a covalent bond. A lewis base is a substance that can donate a pair of electrons to form a covalent bond.

In this reaction, the ammonia is a lewis base. The ammonia will give the extra pair of electrons to the other substance to form an additional bond with it.

RECAP: Arrhenious: Acids make H+ and Bases make OH- Brønsted–Lowry: Acids are proton donors and Bases are proton acceptors Lewis: Acids accepts a pair of electrons, Bases donates a pair of electrons

Concentration: A recap Molarity: moles/L solution –Used in calculations involving pH Molality: moles/ kg solvent –Used in calculations with colligative properties

Concentration: something new! Volume %: (v/v %) –volume of solute / volume of solution x 100 Example: wine is about 12% v/v ethanol. –This means that there are 12 mL ethanol for every 100 mL of wine. Example: rubbing alcohol is sold at 70% v/v. –This could be made by placing 700 mL isopropyl alcohol into a container and adding sufficient water to obtain 1000 mL solution or placing 70 mL isopropyl alcohol into a container and adding enough water to make 100 mL solution

More concentration units: Mass/volume percent: (%m/v): Mass solute (g or kg) / volume of solution (mL or L) x 100 –Use g with mL or kg with L Mass/mass percent: (%m/m): mass of solute / mass of solution x 100 –Same mass unit must be used for solute and solution

Mole fraction This is actually a review! You used this for partial pressure of gasses! Moles of solute / (moles solute + moles solvent)

STOP!

Acid-Base Reactions: AKA Neutralization Reactions Strong acid (H+ ion in front of formula) with Strong Base (hydroxide ion in the back of formula) usually produces water and a salt HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) H 2 SO 4(aq) + KOH (aq)  K 2 SO 4(aq) + H 2 O (l) BALANCE ME!!!

Neutralization reactions These are neutralization reactions – produce neutral solutions and are double replacement reactions.

Titrations A titration is a method of analysis that will determine the precise endpoint of a reaction. It is used to determine the precise quantity of unknown reactant in the titration flask. A buret is used to deliver the second KNOWN reactant to the flask An indicator or pH Meter is used to detect the endpoint of the reaction

Titrations Used to determine the concentration of an unknown acid or base. Done using acid/base indicators (change color as the pH of a solution changes) OR a pH meters.

3 steps to titration problems Step 1: A measured volume of acid or base of unknown concentration is added to a flask Step 2: An indicator is added to the flask that will change the color of the solution when the reaction is complete.

Step 3: A measured volume of acid or base with a known concentration (standard solution AKA: titrant) is added to the flask until the indicator changes color.

Titration vocabulary End point – the point where the indicator just changes color. Equivalence point – midpoint on the pH titration curve. Perfectly reacted: No limiting reagent, no excess reactant left over. Running acid into the base Notice: solution starts with a high pH (basic) and it slowly turns acidic (low pH)

Running base into the acid Notice: solution starts with a low pH (acidic) and it slowly turns basic (high pH)

Titration problems Example 1: How many moles of sulfuric acid are required to neutralize 0.50 moles of sodium hydroxide? 1. Write the equation H 2 SO 4(aq) + 2NaOH (aq)  2H 2 O (l) + Na 2 SO 4(aq) notice: It’s balanced!!!

Titration problems 2. What is given? Mol NaOH = 0.50 mol 3. Mole ratio? 2 moles NaOH 1 mole H 2 SO 4

4. Solve 0.50 mol NaOH 1 mol H 2 SO 4 2 mol NaOH = 0.25 mol H 2 SO 4

Example 2: A 25 mL solution of H 2 SO 4 is completely neutralized by 18 mL of 1.0M NaOH. What is the concentration of H 2 SO 4 ? H 2 SO 4(aq) + 2NaOH (aq) 2H 2 O (l) + Na 2 SO 4(aq) notice: It’s balanced!!! Always balance!

List what you know List what you know: [base] = 1.0M NaOH volume base = 18mL =.018L volume acid = 25mL =.025L Mole ratio: 1mole H 2 SO 4 /2 moles NaOH Find: molarity of the H 2 SO 4

Conversion steps: L NaOH –> mol NaOH –> mol H 2 SO 4 –> [H 2 SO 4 ] =.0090 moles H 2 SO 4.018L NaOH 1.0 mole NaOH1 mole H 2 SO 4 1L NaOH 2 moles NaOH molarity of NaOH solution mole/mole ratio from balanced reaction

This volume came from the original problem. Now solve for molarity: moles H 2 SO 4 /0.025L =.36 M

Tips on solving titration problems Always start your dimensional analysis problem with the volume of the solution you KNOW the concentration of. Always make sure you balance the reaction before you insert the mole-mole ratio