Still Confused?

Try reviewing these fundamentals first

- Home
- GCSE Chemistry
- Acid Base Chemistry and pH

Still Confused?

Try reviewing these fundamentals first

Nope, got it.

That's the last lesson

Start now and get better maths marks!

Get Started NowStart now and get better maths marks!

Get Started NowStart now and get better maths marks!

Get Started NowStart now and get better maths marks!

Get Started Now- Intro Lesson: a3:32
- Intro Lesson: b4:45
- Intro Lesson: c4:14
- Lesson: 1a8:01
- Lesson: 1b2:42

In this lesson, we will learn:

- To recall the equilibrium ionization expressions for
*weak*acids and bases. - How to relate K
_{a}and K_{b}for conjugate acid/base pairs. - How to calculate concentration of aqueous ions in weak acid/base solutions.

- In Autoionization of water, we looked at the equilibrium:
2 H _{2}O_{(l)}$\, \rightleftharpoons \,$ H_{3}O^{+}_{(aq)}+ OH^{-}_{(aq)}

The equilibrium constant was expressed as:K _{w}= [H_{3}O^{+}] [OH^{-}] = 1 * 10^{-14}at 25^{o}C

We saw the effect of adding strong acids/bases to equilibrium concentrations of water and dissolved ions. That was straightforward because__strong acids and bases experience 100% dissociation in water__.

We also saw in - In strong and weak acids and bases that
__weak acids and bases do not experience 100% dissociation__. This makes expressions for their dissociation in water more complicated. - Every weak acid has an
and a weak base a*acid dissociation constant, K*_{a}__base dissociation constant,__. These are equilibrium constants showing how much the acid/base dissociates when dissolved in water (aqueous solution).*K*_{b}

For a weak acid HX dissolved in water:*K*= $\frac{[H_3O^{+}] [X^{-}]}{[HX]}$_{a}

For a weak base B dissolved in water:*K*= $\frac{[HB^{+}] [OH^{-}]}{[B]}$_{b}

- Just like in the ionization of water and other equilibria; this is an equilibrium constant expression. This means that
__the higher the K__(because the concentrations of the dissociated ions, in the numerator, are larger values) and therefore_{a}/K_{b}value, the greater the degree of dissociation__the stronger the acid or base__. - Remember that
__for strong acids and bases K__. This is because in the K_{a}/K_{b}values are not normally used_{a}expression, their [HX] or [B] is equal to or almost zero due to complete dissociation, so the values are incredibly large. __For strong acids, pK__and is more appropriate to use for strong acids, instead of the extremely large K_{a}is used instead of K_{a}. pK_{a}is the negative logarithm of the K_{a}value_{a}values they have.

- Just like in the ionization of water and other equilibria; this is an equilibrium constant expression. This means that
- In Conjugate acids and bases, we learned in a conjugate pair that a stronger conjugate acid will have a weaker the conjugate base. This relationship affects the concentration of aqueous ions and therefore affects the K
_{a}and K_{b}values for a conjugate acid/base pair!

Consider the equations for the conjugate pair acid-base pair HX and X^{-}:Conjugate acid: $\qquad$ HX + H _{2}O $\, \rightleftharpoons \,$ X^{-}+ H_{3}O^{+}$\qquad$*K*= $\frac{[H_3O^{+}] [X^{-}]}{[HX]}$_{a}Conjugate base: $\qquad$ X ^{-}+ H_{2}O $\, \rightleftharpoons \,$ HX + OH^{-}$\qquad$*K*= $\frac{[HX] [OH^{-}]}{[X^{-}]}$_{b}

- Both equations depend on [X
^{-}] and [HX] so K_{a}and K_{b}themselves can be related, and terms cancelled out:

As you can see, the result is the product of [H_{3}O^{+}] and [OH^{-}] which is the expression for K_{w}. Therefore for a conjugate pair:K _{a}* K_{b}= K_{w}

- Both equations depend on [X

- Introduction
__What is the acid/base dissociation constant?__a)From K_{w}to K_{a}and K_{b}.b)The K_{a}and K_{b}expressions.c)Relating K_{a}, K_{b}and K_{w}using conjugate pairs. - 1.
**Use the relationship between the K**_{a}and K_{w}expressions to find an unknown K_{b}value.a)Ethanoic acid, or acetic acid CH_{3}COOH is a weak acid with K_{a}= 1.4*10^{-5.1}- Write the formula of its conjugate base.
- Find the K
_{b}for its conjugate base using the relationship between K_{a}/K_{b}and K_{w}.

b)Explain using the K_{a}/K_{b}expression why K_{a}and K_{b}values are not normally used when studying strong acids and bases.^{1}**Source for K**ATKINS, P. W., & DE PAULA, J. (2006). Atkins' Physical chemistry. Oxford, Oxford University Press._{a}acid dissociation constants:

11.

Acid Base Chemistry and pH

11.1

Introduction to acid-base theory

11.2

Conjugate acids and bases

11.3

Strong and weak acids and bases

11.4

Autoionization of water

11.5

Acid / base dissociation constant (K_{a} and K_{b})

11.6

Relative strengths of acids and bases

11.7

pH, pOH and antilogs

11.8

Mixing strong acids and bases

11.9

Hydrolysis

11.10

K_{a} and K_{b} calculations

11.11

Acid-base titration