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

Question

### 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−]

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

solved 0
2 years 1 Answer 703 views 0