Acids, Bases & pH

Acids, Bases & pH

Types of solutes

sugar

Non-electrolyte -No dissociation into ions, all molecules in solution

no conductivity

Types of solutes

H+

Cl-

Electrolyte -Dissociation intoions in solution.

conductivity

Electrolytes

• Acids, bases, and salts are electrolytes.– HNO3, NaOH, KBr.

Acid-Base Theory - Arrhenius• Arrhenius defined an acid

as follows:– An acid is a substance that

produces H+ ions when dissolved in water (now described as hydronium rather than H+).

Hydronium Ion (H3O+) or (H+(aq))

Acid-Base Theory - Arrhenius• Arrhenius defined a base

as follows:– A base is a substance that

produces OH- ions when dissolved in water

Acid-Base Theory - Arrhenius

• We can demonstrate these definitions as follows:

HNO3(l) H+(aq) + NO3-(aq)

H2O

Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)H2O

pH scale• The pH scale is used to indicate how acidic or

basic a solution is by measuring the relative amount of H+ and OH- in a solution.

The pH scale

pH and pOH scales

Acidic Neutral Basic

pH + pOH = 14.00

pH and pOH formulas

• Remember that H+ = H3O+

1. pH = -log[H+]

2. pOH = -log[OH-]

3. [H+] x [OH-] = 1.0 x 10-14

4. pH + pOH = 14.00

5. [H+] = antilog(-pH)

6. [OH-] = antilog(-pOH)

Find the [OH-] in a solution where [H3O+] = 6.80 x 10-10.

1. pH = -log[H+] 2. pOH = -log[OH-]3. [H+] x [OH-] = 1.0 x 10-14 4. pH + pOH = 14.005. [H+] = antilog(-pH)6. [OH-] = antilog(-pOH)

Find the [OH-] in a solution where [H3O+] = 6.80 x 10-10.

• [H+] x [OH-] = 1.0 x 10-14

• [6.80 x 10-10] x [OH-] = 1.0 x 10-14

1.47 x 10-5 M

Find the pH of a solution where [H3O+] = 7.01 x 10-6.


Find the pH of a solution where [H3O+] = 7.01 x 10-6.

• pH = -log[H+]

• pH = -log[7.01 x 10-6]

5.15

Find the [OH-] in a solution with a pOH of 4.976.


Find the [OH-] in a solution with a pOH of 4.976.

[OH-] = antilog(-pOH)[OH-] = antilog(-4.976)

1.06 x 10-5 M

What is the pH of a solution with a [OH-] = 9.80 x 10-9.


What is the pH of a solution with a [OH-] = 9.80 x 10-9.

pOH = -log[OH-]pOH = -log[9.80 x 10-9]

pOH = 8.01pH + pOH = 14.00

14.00 – 8.01 =

5.99

Acid Base Neutralization

• Acid + Base

• HCl + NaOH • HCl + NaOH NaCl + HOH

• Acid + Base Salt + Water

salt water

Acid-Base Neutralization Reactions

HCl + NaOH NaCl + H2O

• The products of an acid-base neutralization reaction are a salt made up of the positive ion of the base (Na+) and the negative ion (Cl-) of the acid, and water.

HCl + NaOH NaCl + H2O

•Remember, in writing the correct formula for the salt, the total positive charge of the ions in the salt must equal the total negative charge of the ions in the salt.

•Make sure you write the correct formula for the products, salt and water first, then balance the equation.

Write the balanced equation for the acid-base neutralization of perchloric acid, HClO4 and potassium hydroxide, KOH

HClO4 + KOH KClO4 + H2O

Write the balanced equation for the acid-base neutralization of nitric acid, HNO3 and

calcium hydroxide Ca(OH)2

Ca(OH)2 + 2HNO3 Ca(NO3)2 + 2H2O

MA x VA = MB x VB

• Acids and bases often react together in what is called a neutralization reaction. The amount (volume) of acid or base and the strength (molarity) of acid and base needed to neutralize each other can

“sometimes” be found by using the formula above.

MA x VA = MB x VB

• We can use this formula when the acid and base react together in a 1:1 ratio.

Formula can be used:

HClO4 + KOH KClO4 + H2OFormula cannot be used:

Ca(OH)2 + 2HNO3 Ca(NO3)2 + 2H2O

MA x VA = MB x VB

• 15.6ml of acid neutralize 25.8ml of 0.50M base. What is the molarity of the acid?

0.83M

MA x VA = MB x VB

• 32.6 ml of 1.25M acid neutralize how many ml of 5.60M base?

7.28 mL

Homework

• Summarize the Lab: “Acid – Base Titration”

• Worksheet: Acids, Bases and pH

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Acids, Bases & pH