Chapter 20 - Acids, Bases, and pH

Question 1

What is a Bronsted-Lowry acid defined as?

Answer 1

A proton donor

Question 2

What is a Bronsted-Lowry base defined as?

Answer 2

A proton acceptor

Question 3

What is a conjugate acid-base pair?

Answer 3

Two species that can be interconverted by transfer of a proton.

Question 4

In the dissociation of HCl to H+ and Cl-, identify the conjugate acid and base.

Answer 4

HCl releases a proton so is therefore the conjugate acid.

Cl- accepts a proton so is therefore the conjugate base.

Question 5

When water is the base, what is formed?

Answer 5

H(3)O+

Hydronium ion

Question 6

What do the terms monobasic, dibasic and tribasic refer to?

Answer 6

The total number of hydrogen ions in the acid that can be replaced per molecule in an acid-base reaction.

Question 7

How can redox reactions be simplified?

Answer 7

Remove the spectator ions

Question 8

What is the word equation for the reaction of an acid and a metal?

Answer 8

acid + metal -> salt + hydrogen

Question 9

What is the word equation for the reaction of an acid and a carbonate?

Answer 9

acid + carbonate -> salt + water + carbon dioxide

Question 10

What is the word equation for the reaction of an acid and a base?

Answer 10

acid + base -> salt + water

Question 11

What is the word equation for the reaction of an acid and an alkali?

Answer 11

acid + alkali -> salt + water

Question 12

What is the relationship between pH and concentration of H+?

Answer 12

Low value [H+] = high pH

High value [H+] = low pH

Question 13

What is the equation for working out pH from [H+]?

Answer 13

pH = -log[H+] (base 10)

Question 14

How is the pH calculated for a strong acid?

Answer 14

Assume it fully dissociates

Therefore [H+] = [HA]

Question 15

How is the new pH calculated for a strong acid on dilution?

Answer 15

Work out change in concentration of HA and therefore [H+]

Then put back into pH = -log[H+]

Question 16

How is the pH calculated for a weak acid?

Answer 16

Ka = [H+][A-]/[HA]
Ka[HA] = [H+][A-]
[H+] = [A-]
[H+]^2 = Ka[HA]
[H+] = sqrt(Ka[HA])

Question 17

What is Kw?

Answer 17

The ionic product of water

Question 18

What is the value of Kw?

Answer 18

1x10^-14 mol2dm-3 at 298K

given on data sheet

Question 19

How can pH of strong bases be found when given the concentration of OH-?

Answer 19

Kw =[H+][OH-]=1x10^14

Find [H+]
Then pH=-log[H+]

Question 20

How can pH of weak bases be found when given the concentration of OH-?

Answer 20

Similar method to weak acids, with the additional step of Kw = [H+][OH-]

Question 21

what is the pOH formula

Answer 21

pOH= -log[OH(aq)]

Question 22

What is the formula of [OH-(aq)]

Answer 22

10^-pOH

Question 23

pKa value formula

Answer 23

pKa = -log ka

Question 24

What is the unlog of pKa

Answer 24

Ka = 10^-pKa

Question 25

limitations of using approximations to Ka related calculations for ‘stronger’ weak acids

Answer 25

[HA]equilibrium ~ [HA]undissociated may no longer be valid.

The approximation that the concentration of acid is much greater than the H+ concentration at equilibrium is not valid