Weak acids are compounds that partially dissociate to produce an equilibrium concentration of H

CH_{3}COOH + H_{2}O <--> CH_{3}COO^{-}
+ H_{3}O^{+}

Weak bases partially react with water to produce an equilibrium
concentration of OH^{-}. An example equilibrium is ammonia in
water:

NH_{3} + H_{2}O <--> NH_{4}^{+}
+ OH^{-}

What is the pH of a 0.200 M solution of acetic acid
(K_{a} = 1.75E-5)

These are the important equations:

CH_{3}COOH + H_{2}O <--> CH_{3}COO^{-}
+ H_{3}O^{+}

( [H_{3}O^{+}] [CH_{3}COO^{-}]
) / [CH_{3}COOH] = 1.75 x 10^{-5}

Searching for the [H_{3}O^{+}]

[H_{3}O^{+}] = [CH_{3}COO^{-}]
= x

The [CH_{3}COOH] started at 0.2 M and went
down as CH_{3}COOH molecules dissociated.

0.2 - x

We now have our equation:

1.75E-5 = x^{2} / (0.2 - x)

-x^{2} - 1.75E-5x + 3.5E-6 = 0

Solving this equation, we get:

x = 0.00186 M

pH -log 0.00186 = 2.73

What is the pH of a 0.200 M solution of acetic acid
(K_{a} = 1.75E-5)

These are the important equations:

CH_{3}COOH + H_{2}O <--> CH_{3}COO^{-}
+ H_{3}O^{+}

( [H_{3}O^{+}] [CH_{3}COO^{-}]
) / [CH_{3}COOH] = 1.75 x 10^{-5}

Searching for the [H_{3}O^{+}]

[H_{3}O^{+}] = [CH_{3}COO^{-}]
= x

The [CH_{3}COOH] started at 0.2 M and went
down as CH_{3}COOH molecules dissociated. In this approximation
we
simplify by ignoring that molecules dissociate from the original conc.
substituting (0.2-x) by 0.2, since x is rather small.

This new equation is easy to solve:

1.75E-5 = x^{2} / 0.2

(x^{2})^{1/2} = 3.5E-6^{1/2}

Solving this equation, we get:

x = 0.00187 M

pH = -log 0.00187 = 2.73

1) Insert equation of dissociation: CH

2) Insert 1.75E-5 in K-field

3) Insert 0.2 in CH

4) Calculate

CHEMIX
School - Weak acid base

1) Insert equation of dissociation: CH

2) Insert 1.75E-5 in K-field

3) Insert 0.1 in CH3COOH (Before dissoc.)-field

4) Calculate

.

1) Insert equation of dissociation: CH

2) Insert 1.75E-5 in K-field

3) Insert 0.002 in H

4) Calculate

1) Insert equation of dissociation: CH

2) Insert 1.75E-5 in K-field

3) Insert 4 i pH-field

4) Calculate

1) Insert equation of dissociation: CH

2) Insert 1.75E-5 in K-field

3) Insert 0.1 in CH

4) Insert 0.1 in CH

5) Calculate.

1) Insert equation of dissociation: CH

2) Insert 1.75E-5 in K-field

3) Insert 0.001 in CH

4) Insert 7.0 pH-field

5) Calculate

1) Insert equation of dissociation: HCN + H

2) Insert 5.85E-10 in K-field

3) Insert 1E-6 in H

4) Insert 0.2 i HCN (before dissoc.) field.

5) Calculate

Ex.2: 0.2286M

Ex.3: Dissoc=0.0001M, un-dissoc=0.000571M

Ex.4: pH=4.757

Ex.5: 0.1745 mol

Ex.6: 0.000116 mol

Acid base indicators chart

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