## Weak acid/base

Weak acids are compounds that partially dissociate to produce an equilibrium concentration of H+. An example equilibrium is acetic acid in water :

CH3COOH + H2O <--> CH3COO- + H3O+

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

NH3 + H2O <--> NH4+ + OH-

### Sample Weak Acid Problem (Exact Solution)

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

These are the important equations:

CH3COOH + H2O <--> CH3COO- + H3O+

( [H3O+] [CH3COO-] ) / [CH3COOH] = 1.75 x 10-5

Searching for the [H3O+]

[H3O+] = [CH3COO-] = x

The [CH3COOH] started at 0.2 M and went down as CH3COOH molecules dissociated.

0.2 - x

We now have our equation:

1.75E-5 = x2 / (0.2 - x)

-x2 - 1.75E-5x + 3.5E-6 = 0

Solving this equation, we get:

x = 0.00186 M

pH -log 0.00186 = 2.73

### Sample Weak Acid Problem (Approximation)

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

These are the important equations:

CH3COOH + H2O <--> CH3COO- + H3O+

( [H3O+] [CH3COO-] ) / [CH3COOH] = 1.75 x 10-5

Searching for the [H3O+]

[H3O+] = [CH3COO-] = x

The [CH3COOH] started at 0.2 M and went down as CH3COOH 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 = x2 / 0.2

(x2)1/2 = 3.5E-61/2

Solving this equation, we get:

x = 0.00187 M

pH = -log 0.00187 = 2.73

### Sample Weak Acid Problem (CHEMIX Solution)

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

1) Insert equation of dissociation: CH3COOH + H2O > CH3COO- + H3O+
2) Insert 1.75E-5 in K-field
3) Insert 0.2 in CH3COOH (Before dissoc.)-field
4) Calculate

CHEMIX School - Weak acid base ### Examples from program - problems and solutions

Ex. 1: Calculate pH and H3O+-conc. in a 0.1M solution of acetic acid(Ka=1.75E-5).
Solution:
1) Insert equation of dissociation: CH3COOH + H2O > CH3COO- + H3O+
2) Insert 1.75E-5 in K-field
3) Insert 0.1 in CH3COOH (Before dissoc.)-field
4) Calculate

Ex. 2: Decide the amount of undissociated acetic acid in a [H3O+] = 0.002M solution (Ka=1.75E-5)
. Solution:
1) Insert equation of dissociation: CH3COOH + H2O > CH3COO- + H3O+
2) Insert 1.75E-5 in K-field
3) Insert 0.002 in H3O+/OH--field
4) Calculate

Ex. 3:Decide conc. of dissoc. and undissoc. acetic acid in a pH=4 solution (Ka=1.75E-5)
Solution:
1) Insert equation of dissociation: CH3COOH + H2O > CH3COO- + H3O+
2) Insert 1.75E-5 in K-field
3) Insert 4 i pH-field
4) Calculate

Ex. 4:Find pH in a solution containing 0.1M CH3COONa and 0.1M acetic acid(Ka=1.75E-5)
Solution:
1) Insert equation of dissociation: CH3COOH + H2O > CH3COO- + H3O+
2) Insert 1.75E-5 in K-field
3) Insert 0.1 in CH3COOH (Before dissoc.)-field
4) Insert 0.1 in CH3COO- common ion field
5) Calculate.

Ex. 5:a)How many moles of NaAc (CH3COONa) must be added in a 1 dm3 0.001M CH3COOH-solution making a pH=7.0-buffer (Ka=1.75E-5).
Solution:
1) Insert equation of dissociation: CH3COOH + H2O > CH3COO- + H3O+
2) Insert 1.75E-5 in K-field
3) Insert 0.001 in CH3COOH (Before dissoc.)-field
4) Insert 7.0 pH-field
5) Calculate

Ex. 6:How many moles of NaCN must be added in a 1 dm3 0.2M HCN before [H3O+] = 1.0E-6 ? (Ka=5.85E-10)
Solution:
1) Insert equation of dissociation: HCN + H2O > CN- + H3O+
2) Insert 5.85E-10 in K-field
3) Insert 1E-6 in H3O+-field
4) Insert 0.2 i HCN (before dissoc.) field.
5) Calculate

### Solutions

Ex.1: pH=2.88, H3O+-conc.=0.001314M
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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