More reactions of Acids & Bases

Neutralization Reaction = acid + base acid + base salt + salt + water water – The salt comes from the cation of the base & the anion

of the acid. – Water is made from the H+ on the acid & the OH- on

the base. Molecules with OH- can be acidic or amphotericamphoteric

depending on the other oxygens present. More oxygens take electrons away from the O-H

bond and make it easier to break (more polar/more acidic)

Water

Water is polar--it has a positivepositive and a negativenegative end. – Water can form OH- and H+

– H+ reacts with water to form H3O+ (hydronium ion)

– This can be written as the self ionization of water: 2H2H22O O H H33OO++ + OH + OH--

More on Water

In pure water only 1 x 10-7 M of H+ and OH- form. Pure water is neutral.

You can multiply the concentration of H+/H3O+ and OH- to get the ion product constant for water Kw.

KKww= 1 x 10= 1 x 10–14–14 M M22 [H[H++] [OH] [OH-1-1] = K] = Kww

[H[H+ + or H or H33OO + +] [OH] [OH-1-1] = 1 x 10] = 1 x 10–14–14 M M22 From this equation, it is possible to figure out

either [H+1] or [OH-1] if the other one is known.

Sample Problem

If the [OH-] in a solution is 1.15 x 10-12 M, what is the [H3O+] of this solution?

[H[H33OO++][OH][OH--]= 1 x 10]= 1 x 10-14-14 M M22

[[HH33OO++] = ] = KKww

[OH[OH--] ]

[[HH33OO++] = ] = 1 x 101 x 10-14-14 M M22 = 0.00870 M = 0.00870 M

1.15 x 101.15 x 10-12-12 M M

Your Turn!

If the [H3O+] in a solution is 2.98 x 10-5 M, what is the [OH-] of this solution?

[OH[OH--] = ] = KKww

[[HH33OO++] ]

[OH[OH--] = ] = 1 x 101 x 10-14-14 M M22 = 3.36 x 10= 3.36 x 10-10-10 M M 2.98 x 102.98 x 10-5-5 M M

Definitions An acidic solution is one in which the [H3O+] is greatergreater than the

[OH-]. An basic solution is one in which the [OH-] is greater than the [H3O+]. pHpH was developed to be an easier way to calculate acid & base

concentration than Kw. The pH of a solution is the negative logarithm of the hydrogen ion

concentration.pH = -log[HpH = -log[H++] or pH = -log[H] or pH = -log[H33OO++]]

The pOH of a solution is the negative logarithm of the hydroxide ion concentration.

pOH = -log[OHpOH = -log[OH--]]

pH and pOH

In a neutral solution, [H3O+] = 1 x 10-7 M. The pH of the solution is 7.07.0

The pH of pure water or a neutral solution is 7.0

A neutral solution has a pOHpOH of 7.0 Simple Relationship:

pH + pOH = 14pH + pOH = 14

Sample Problem What is the pH of a HBr solution with a

hydrogen-ion concentration of 1.36 x 10-4 M? pOH?pH = -log[pH = -log[HH33OO++]]

= -log(1.36 x 10= -log(1.36 x 10-4-4 ) ) = -(-3.87)= -(-3.87) = 3.87= 3.87

pH + pOH = 14pH + pOH = 14pOH = 14 – 3.87 = 10.13pOH = 14 – 3.87 = 10.13

Your Problem What is the pH of a HCl solution with a

hydrogen-ion concentration of 9.85 x 10-2 M? pOH?pH = -log[pH = -log[HH33OO++]]

= -log(9.85 x 10= -log(9.85 x 10-2-2))

= 1.01= 1.01

pOH = 14 – 1.01 = 12.99pOH = 14 – 1.01 = 12.99

Try another one! What is the pOH of a NaOH solution with a

hydroxide-ion concentration of 6.35 x 10-3 M? pH? Is this an acidic, basic or neutral solution?pOH = -log[OHpOH = -log[OH--]]

= -log(6.35 x 10= -log(6.35 x 10-3-3)) = -(-2.20)= -(-2.20) = 2.20= 2.20

pH = 14 – 2.20 = 11.80pH = 14 – 2.20 = 11.80pH = 11.80 = BasicpH = 11.80 = Basic

Another one What is the pH of a KOH solution with a

hydroxide-ion concentration of 3.45 x 10-4 M?pOH = -log[OH-]pOH = -log[OH-]

= -log(3.45 x 10= -log(3.45 x 10-4-4 ) )

= 3.46= 3.46pH = 14 - pOH = 14 – 3.46pH = 14 - pOH = 14 – 3.46pH = 10.54pH = 10.54

Almost Done! What is the [OH-] and [H+] concentration of an HI

solution with a pH of 4.5?pH = -log[pH = -log[HH33OO++]]

log[log[HH33OO++] = -pH] = -pH

pH = 4.5pH = 4.5

log[log[HH33OO++] = -4.5] = -4.5

[[HH33OO++] = antilog - 4.5 or 10] = antilog - 4.5 or 10-4.5-4.5

[[HH33OO++] =3.16 x 10] =3.16 x 10-5-5 M M[OH[OH--] = ] = 1 x 101 x 10-14-14 M M22 = 3.16 x 10= 3.16 x 10-10-10 M M

3.16 x 103.16 x 10-5-5 M M

Last one! What is the [OH-] and [H3O+] concentration of a

LiOH solution with a pOH of 3.6?pOH = -log[OHpOH = -log[OH--]]log[OHlog[OH--] = -pOH] = -pOHpOH = 3.6pOH = 3.6log[OHlog[OH--] = -3.6] = -3.6

[OH[OH--] = antilog – 3.6 or 10] = antilog – 3.6 or 10-3.6-3.6

[OH[OH--] = 2.51 x 10] = 2.51 x 10-4-4 M M [[HH33OO++] = ] = 1 x 101 x 10-14-14 M M22 = 3.98 x 10= 3.98 x 10-11-11 M M

2.51 x 102.51 x 10-4-4 M M

What 2 teams are tied for the most NCAA basketball final four appearances?

Providence College (my school) has 1 in 1987.

UCLA and North Carolina with 18 each