Acids and Bases

Properties of Acids and Bases

Pg 236

• Why do Acids and bases change ACIDS BASES

Sour taste (vinegar) Bitter taste (baking soda)

React with some metal to form H2 gas (Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

Feels slippery (soap)

Turns blue litmus red Turns red litmus blue

Arrhenius Definition

• Acid: produces H+ (or H3O+) when dissolved in water

HCl(aq) H+(aq) + Cl-(aq)H+(aq) + H2O(l) H3O+(aq)

NOTE: H3O+ = hydronium ion

• Base: produces OH- when dissolved in water.

NaOH(s) Na+(aq) + OH-(aq)

Bronsted-Lowry Definition Acids: proton (H+) donors

HF(aq) H+(aq) + F-(aq)H+(aq) + H2O(l) H3O+(aq)

Bases: proton acceptorsNH3(aq) + H2O(l) NH4

+(aq) + OH-(aq)

H2O: acts as an acid and a base = AMPHOTERIC

H

NH

H

H+ N+

H

H H

H

Strength of AcidsStrong Acids: ionize (splits up into ions) almost 100% in

water mostly ions in solution amount of HCl present is negligeable

HCl(aq) H+(aq) + Cl-(aq)

Weak acids: ionize poorly in water not many of these ions present in solution mostly acetic acid (HC2H3O2)

HC2H3O2(aq) C2H3O2-(aq) + H+(aq)

NOTE: strong acids are strong electrolytes and will conductelectricity better than weak acids.

Strength of BasesStrong Bases: ionize almost 100% in water

NaOH(s) Na+(aq) + OH-(aq)

Weak Bases: ionize poorly in water

NH3(l) + H2O(l) NH4+(aq) + OH-(aq)

NOTE: strong bases are strong electrolytes

Conjugate Acids and Conjugate Bases

• these differ by only one proton• Examples

HCl Cl-

SO42- HSO4

-

Lose a proton

Gain a proton

Acid Conjugate base of HCl

Base Conjugate acid of SO4

2-

Reactions with Water

Conjugate acid-base pair: CH3CO2H/CH3CO2-

Conjugate acid-base pair: H2O/H3O+

H C

H

H

C

O

O H

+ O

H H

H C

H

H

C

O

O - O

+ H H

H

acetic acid water acetate ion hydronium ion

Monoprotic, Diprotic and TriproticMonoprotic

·        donates one acidic proton·        eg: HCl + H2O H3O+ + Cl-

·        only one H+ to donate

Diprotic

·        donates two acidic protons·        eg: H2SO4 + H2O H3O+ + HSO4

-

·        HSO4- + H2O H3O+ + SO4

2-

Triprotic

·        donates three acidic protons·        eg: H3PO4 + H2O H3O+ + H2PO4

-

·        three H+ to donate

Homework

• Pg 251 #1, 2• Pg 253 #4, 5, 6

pH < 7 acidicpH = 7 neutralpH > 7 basic

a measure of acid strengthBy definition all acids contain at least one acidic proton

= H+

HA is a symbol used to represent any general acid HA H+(aq) + A-(aq)

H+ + H2O H3O+

[H+] = [H3O+]

If a lot of H3O+ is produced the solution is very acidic.

pH is directly related to [H3O+].

pH

H2O(l) + H2O(l) H3O+(aq) + OH-(aq)

This reaction does not occur to any great extent.[H3O+] = 1 x 10-7 mol/L[OH-] = 1 x 10-7 mol/LBecause both concentrations are equal water is said to be neutral.Therefore, if [H3O+] = [OH-] neutral [H3O+] > [OH-] acidic

[H3O+] < [OH-] basic

NOTE: [H3O+][OH-] = 1.0 x 10-14

Self-Ionization of Water

• Expressing hydronium concentrations in scientific notation isn’t very convenient. The pH scale was developed to make the expression of H3O+ concentration more convenient.

• [H3O+] is the concentration in mol/L

pH = -log[H3O+]

The concentration of H3O+ is 1.0 x 10-7. Calculate the pH.

Example 1: pH of Water

pH = -log[H3O+]

= -log(1.0 x 10-7)

= -(-7)

= 7

The pH of water is 7. Therefore pH 7 is neutral.

Determine the pH of a 1M solution of HCl.

Example 2

HCl (aq) H+ + Cl-

1M x

Therefore [H3O+] = 1

Therefore, pH = -log[H3O+]

= -log(1)

= 0

Therefore a 1M solution of HCl has pH 0.

What is the pH of a 0.01M solution of HCl?

Example 3

HCl (aq) H+ + Cl-

[H3O+] = 0.01 M

Therefore, pH = -log[H3O+]

= -log(0.01)

= 2

What is the pH of a 1M NaOH solution?

Example 4

pOH = -log[OH-] = -log(1) = 0

pH + pOH = 14

pH = 14 – pOH

pH = 14

Therefore a 1M solution of NaOH has pH 14. The pH

of a very basic solution.

Determine the pH of a 0.01M NaOH solution.

Example 5

pOH = -log[OH-] = -log(0.01) = 2

pH + pOH = 14 pH = 14 – pOH pH = 12

Therefore a 1M solution of NaOH has pH 12. The pH

of a basic solution.

The pH reading of a solution is 10.33. What is its hydrogen ion concentration?

Example 6

10-pH = [H+]

10-10.33 = [H+]

4.7 10-11 mol/L = [H+]

Base ten logarithm represents

an exponent log10(100) =2

102

Calculate the pH of a 0.00242 M H2SO4 solution .

Example 7

H2SO4 2H+ + SO42-

0.00242 M 0.00484M

pH = - log[H+]

= -log(0.00484)

= -(-2.315)

= 2.32

Homework

• Pg 239 #1, 2• Pg 242 #5, 7, 9, 10

Acid-Base Indicators

• Can determine if the solution is acidic, basic or natural using various indicators

• Litmus paper, bromothyomol blue, phenolphthalein are some examples.

• Depending on the indicator they change colour at varying pH levels.

• Need to use various indicators to solve the pH level

Chart

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