## Complete this table of values for four aqueous solutions at 25 °c.

### Question:

Complete this table of values for four aqueous solutions at 25 °c.

Solution A: H^+ = 7.3 * 10^-13, OH^- = ?, pH = ?, pOH = ?

Solution B: H^+ = ?, OH^- = 0.078, pH = ?, pOH = ?

Solution C: H^+ = ?, OH^- = ?, pH = 5.89, pOH = ?

Solution D: H^+ = ?, OH^- = ?, pH = ?, pOH = 11.99

**Calculating for pH, pOH, [H+] and [OH−]**

**The autoionization for water is given by**

[H+][OH−]=Kw

where Kw=1.0×10−14

The pH-pOH relation is given by

pH+pOH=14

**The relationship between pH and hydrogen ion concentration is given by**

[H+]=10−pH

and

pH=−log[H+]

**While the relationship between pOH and the hydroxide ion concentration is given by**

[OH−]=10−pOH

and

pOH=−log[OH−]

### Answer & Explanation:

**Solution A**

[H+]=7.3×10−13

Solving for [OH−]

[OH−]=1.0×10−14/[H+]

[OH−]=1.0×10−14/7.3×10−13

[OH−]=1.4×10−2

**Solving for pH**

pH=−log[H+]

pH=−log7.3×10−13

pH=12.1

**Solving for pOH**

pH+pOH=14

pOH=14−pH

pOH=14−12.1

pOH=1.9

**Solution B**

[OH−]=0.078

**Solving for [H+]**

[H+]=1.0×10−14/[OH−]

[H+]=1.0×10−14/0.078

[H+]=1.28×10−13

**Solving for pH**

pH=−log[H+]

pH=−log1.28×10−13

pH=12.89

**Solving for pOH**

pH+pOH=14

pOH=14−pH

pOH=14−12.89

pOH=1.11

**Solution C**

pH=5.89

**Solving for pOH**

pH+pOH=14

pOH=14−pH

pOH=14−5.89

pOH=8.11

**Solving for [H+]**

[H+]=10−pH

[H+]=10−5.89

[H+]=1.29×10−6

**Solving for [OH−]**

[OH−]=1.0×10−14/[H+]

[OH−]=1.0×10−14/1.29×10−6

[OH−]=7.75×10−9

**Solution D**

pOH=11.99

**Solving for pH**

pH+pOH=14

pH=14−pOH

pH=14−11.99

pH=2.01

**Solving for [H+]**

[H+]=10−pH

[H+]=10−2.01

[H+]=9.77×10−3

**Solving for [OH−]**

[OH−]=1.0×10−14/[H+]

[OH−]=1.0×10−14/9.77×10−3

[OH−]=1.02×10−12

## Answer ( 1 )

Answer Above.