8.6 Calculations Involving Acids And Bases Flashcards Preview

Chemistry Chapter 8 - Acids and Bases > 8.6 Calculations Involving Acids And Bases > Flashcards

Flashcards in 8.6 Calculations Involving Acids And Bases Deck (15):
1

What information do the acid and base dissociation constants give?

The strength of acids and bases.

2

Using the equation
HA (aq) H+ (aq) + A- (aq)
what is the equation for Ka?

Ka = [A-(aq)] [H+(aq)] / [HA(aq)]

3

How are Ka and pKa related?

pKa = -logKa
Ka = 10^(-pKa)

4

What does a larger value for Ka indicate?

A stronger acid.

5

What does a smaller value for pKa indicate?

A stronger acid.

6

Using the equation
B (aq) + H2O (l) BH+ (aq) + OH- (aq)
what is the equation for Kb?

Kb = [BH+(aq)] [OH-(aq)] / [B(aq)]

7

How are Kb and pKb related?

pKb = -logKb
Kb = 10^(-pKb)

8

What does a larger value for Kb indicate?

A stronger base.

9

What does a smaller value for pKb indicate?

A stronger base.

10

What are the relationship between Ka, Kb and pKa and pKb?

Ka x Kb = Kw
pKa + pKb = pKw
at 25 degrees Celsius pKa + pKb = 14

11

How can pOH be used to simplify the calculations of the pH of a weak base?

pOH is basically the same as pH except it refers to the concentration of OH- ions rather than H+ ions.

12

What are the equations of pOH?

pOH = -log[OH-(aq)]
[OH-(aq)] = 10^(-pOH)

13

How can the ionic product constant expression be converted?

Kw = [H+(aq)][OH-(aq)] can be converted to pKw = pH + pOH or at 25 degrees Celsius pH + pOH = 14

14

How can you calculate the pH of a 0.1000 moldm-3 solution of ammonia with pKb of 4.75?

Use the expression for Kb to calculate [OH-(aq)]
Calculate pOH
Use pOH + pH = 14 to work out pH.

15

How does Kw vary with temperature?

The ionisation of water is endothermic and therefore the degree of ionisation increases as temperature increase. This means that the pH of water decreases s temperature increases (higher contraction of H+ (q) at higher temperature).