Ka Flashcards

1
Q

What is the strength of an acid a measure of?

A

The strength of an acid HA is the extent of its dissociation into H* and A- ions.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

What is a strong acid?

A

A strong acid is an acid that completely dissociates in solution.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

What is Ka?

A

The actual extent of acid dissociation is measured by an equilibrium constant called the acid dissociation constant Ka.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Define the acid dissociation constant.

A

The acid dissociation constant,Ka, of an acid HA is defined as:

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

What are the units of Ka?

A

The units of Ka are always mol dm^-3.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

What does a large value of Ka show?

A

A large Ka value indicates a large extent of dissociation - the acid is strong.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

What is the reason to have to pKa?

A

Values of Ka can be made more manageable if expressed in a logarithmic form called pKa.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Mathematically : pKa =

A

pKa = -log Ka

Ka = 10-pKa

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

What is the Ka expression for a weak acid [HA] explain how its derived.

A

Ka = [H(aq)]^2/[HA(aq)]-[H(aq)]

When HA molecules dissociate, [H*] and [A-] ions are formed in equal quantities.

Because relatively few HA molecules have dissociated, [HA] will have reduced slightly. The equilibrium concentration of HA(aq) will be [HA]-[H*].

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

What is the Ka expression for a weak acid [HA] when making an approximation.

A

Only a very small proportion of HA dissociates. We can assume that the equilibrium concentration of HA will be very nearly the same as the concentration of undissociated HA(aq).

The Ka expression approximates to Ka = [H*(aq)]^2/[HA(aq)]

How well did you know this?
1
Not at all
2
3
4
5
Perfectly