Problem 101P

Complete the table. (The first row is completed for you.)

[H3O+] |
[OH−] |
pOH |
pH |
Acidic or Basic |

1.0 × 10−4 |
1.0 × 10−10 |
10.00 |
4.00 |
Acidic |

5.5 × 10−3 |
________ |
________ |
________ |
________ |

________ |
3.2 × 10−6 |
________ |
________ |
________ |

4.8 × 10−9 |
________ |
________ |
________ |
________ |

________ |
________ |
________ |
7.55 |
________ |

Solution 101P: Here, we are going to complete the given table.

Step1:

pH is the negative logarithm to base 10 of hydrogen ion concentration whereas pOH is the negative logarithm to base 10 of hydroxyl ion concentration, i.e.,

pOH = - log[OH- ] and pH = - log[H+ ]

And pH + pOH =14

We can distinguish acidic, neutral and basic aqueous solutions by the relative values of their pH

Acidic: pH < 7

Neutral: pH = 7

Basic: pH > 7

Step2:

[H3O+] |
[OH-] |
pOH |
pH |
Acidic or Basic |

1.0 x 10-4 |
1.0 x 10-10 |
10.00 |
4.00 |
Acidic |

5.5 x 10-3 |
1.82 x 10-12 |
11.74 |
2.26 |
Acidic |

3.13 x 10-9 |
3.2 x 10-6 |
5.495 |
8.505 |
Basic |

4.8 x 10-9 |
2.08 x 10-6 |
5.681 |
8.319 |
Basic |

2.82 x 10-8 |
3.55 x 10-7 |
6.45 |
7.55 |
Basic |

- Given, [H3O+] = 5.5 x 10-3

Therefore, pH = - log[H3O+ ]

= - log (5.5 x 10-3)

= - log 5.5 - log 10-3

= - 0.740 + 3

= 2.26

Therefore, pOH = 14 - pH

= 14 - 2.26

= 11.74

pOH = - log[OH- ]

Therefore, [OH- ] = 10-pOH

= 10-11.74

= 1.82 x 10-12

Since, pH < 7, therefore the solution is acidic.

b) Given, [OH-] = 3.2 x 10-6

Therefore, pOH = - log[OH-]

= - log (3.2 x 10-6)

= - log 3.2 - log 10-6

= - 0.505 + 6

= 5.495

Therefore, pH = 14 - pOH

= 14 - 5.495

= 8.505

pH = - log[H3O+]

Therefore, [H3O+] = 10-pH

= 10-8.505

= 3.13 x 10-9

Since, pH > 7, therefore the solution is basic.

c) Given, [H3O+] = 4.8 x 10-9

Therefore, pH = - log[H3O+ ]

= - log (4.8 x 10-9)

= - log 4.8 - log 10-9

= - 0.681 + 9

= 8.319

Therefore, pOH = 14 - pH

= 14 - 8.319

= 5.681

pOH = - log[OH- ]

Therefore, [OH- ] = 10-pOH

= 10-5.681

= 2.08 x 10-6

Since, pH > 7, therefore the solution is basic.

d) Given, pH = 7.55

pH = - log[H3O+]

Therefore, [H3O+] = 10-pH

= 10-7.55

= 2.82 x 10-8

pOH = 14 - pH

= 14 - 7.55

= 6.45

pOH = - log[OH- ]