Exercises on pH and pOH

The acidity or basicity of a solution is measured using the logarithmic scales of pH and pOH.

The values ​​of these scales range from 0 to 14 and were obtained from the water autoionization reaction.

Calculations involving the pH of solutions are very common in Enem and vestibular.

With that in mind, we created this list with 10 questions to test your knowledge of the topic.

Also use the resolution comments to learn tips and see step-by-step how to solve the exercises.

General inquiries

1. (Fuvest) The self-ionization of water is an endothermic reaction. A student measured the pH of freshly distilled, CO-free water.₂ and at 50 °C, finding the value 6.6. Suspicious that the measuring device was defective, as he expected the value of 7.0, he consulted a colleague who made the following statements:

(I) its value (6.6) may be correct, as 7.0 is the pH of pure water, but at 25 °C;
(II) the application of Le Chatelier's principle to the balance of water ionization justifies that, with increasing temperature, the concentration of H

⁺ ;
(III) in water, the pH is lower the higher the concentration of H⁺ .

What is stated is correct

a) only in I.
b) only in II.
c) only in III.
d) only in I and II.
e) in I, II and III.

See Answer

Correct alternative: e) in I, II and III.

(I). CORRECT The pH of water is obtained from the ionic product, which is [H₃O⁺].[OH^-].

This expression comes from the equilibrium constant which varies with temperature.

(II). CORRECT The autoionization of water is given by

As the direct reaction is endothermic, it consumes heat and, therefore, the increase in temperature favors the formation of hydronium ions.

By increasing the concentration of H⁺ in solution, there is a decrease in pH.

(III). CORRECT The expression of pH is: pH = - log [H⁺]. As the value is expressed in cologarithm, the higher the concentration of H⁺ lower the pH.

2. (Unicap) Tick column I for correct and column II for wrong.

I - II
0 - 0 The pH of pure water is 7, at 25 °C.
1 - 1 The pH of a 0.01 mol/L hydrochloric acid solution is equal to 2.
2 - 2 The pH of a 0.01 mol/L sodium hydroxide solution is equal to 11.
3 - 3 The pOH of a 0.01 mol/L potassium hydroxide solution is equal to 2.
4 - 4 The higher the pOH of an aqueous solution, the lower the concentration of hydroxyl ions and the higher its acidity.

See Answer

Correct answer: 0.1; 1.I; 2.II; 3I; 4.I.

0.I CORRECT.

The pH of water is given from the ionic product of self-ionization and this value varies according to temperature.

At 25°C the ionic product of the self-ionization of water is 10^-14.

From this value, we calculate the pH.

1.I CORRECT.

2.II WRONG.

3.I CORRECT.

4.1 CORRECT.

A high pOH such as 12 has a low concentration of hydroxyl ions, since [OH^-] = 10^-pOH and greater is its acidity, because:

3. (Fuvest) Among the liquids in the table below:

Liquid	[H^+] mol/L	[oh^-] mol/L
milk	1,0. 10^-7	1,0. 10^-7
sea ​​water	1,0. 10^-8	1,0. 10^-6
Coke	1,0. 10^-3	1,0. 10^-11
prepared coffee	1,0. 10^-5	1,0. 10^-9
teardrop	1,0. 10^-7	1,0. 10^-7
washer water	1,0. 10^-12	1,0. 10^-2

it has only acidic character:

a) the milk and the tear.
b) washer water.
c) prepared coffee and cola.
d) sea water and washer water.
e) Coca-Cola.

See Answer

Correct alternative: c) prepared coffee and cola.

Liquid	pH = - log [H^+]	pOH = -log[OH^-]	Character
milk	7	7	Neutral
sea ​​water	8	6	Basic
Coke	3	11	Acid
prepared coffee	5	9	Acid
teardrop	7	7	Neutral
washer water	12	2	Basic

a) WRONG. Milk and tear are neutral liquids.

b) WRONG. Washer water has a basic character.

c) CORRECT. Acid solutions have:

[H3O^+] > 1,0. 10^-7 mol/L	pH
[oh^-] -7 mol/L	pOH > 7

d) WRONG. Sea water and washer water have a basic character.

e) WRONG. Not only is Coca-Cola an acidic character, but also prepared coffee.

Calculation of pH of solutions

4. (UFRGS) If 90 ml of water are added to 10 mL of an aqueous solution of pH = 4.0, the pH of the resulting solution will be equal to:

a) 0.4
b) 3.0
c) 4.0
d) 5.0
e) 5.5

See Answer

Correct alternative: d) 5.0

1st step: calculate the concentration of H ions⁺ at pH = 4.

2nd step: calculate the number of moles present in 10 mL of solution.

3rd step: calculate the final volume of the solution.

4th step: calculate the molar concentration of the solution.

5th step: calculate the pH of the resulting solution.

5. (UFV) Consider a beaker containing 1.0 L of a 0.20 mol/L solution of hydrochloric acid (HCℓ). To this solution 4.0 g of solid sodium hydroxide (NaOH) were added, stirring until its complete dissolution. Since no significant volume change occurred and the experiment was carried out at 25 °C, tick the correct alternative.

a) The resulting solution will be neutral and will have a pH equal to 7.
b) The resulting solution will be basic and will have a pH equal to 13.
c) The resulting solution will be acidic and will have a pH equal to 2.
d) The resulting solution will be acidic and will have a pH equal to 1.
e) The resulting solution will be basic and will have a pH equal to 12.

See Answer

Correct alternative: d) The resulting solution will be acidic and will have a pH equal to 1.

1st step: calculate the molar mass of NaOH.

2nd step: calculate the number of moles of NaOH.

3rd step: calculate how much acid has reacted with the base.

HCl and NaOH react to form salt and water in a neutralization reaction.

As the reaction stoichiometry is 1:1, we have: 0.1 mol of hydrochloric acid reacted with 0.1 mol of sodium hydroxide.

However, the initial solution contained 0.2 mol of HCl and after the reaction with NaOH only 0.1 mol remained, changing the pH of the solution.

4th step: calculate the new pH of the solution.

Calculation of pOH of solutions

6. (Vunesp) At 25 °C, the pOH of a hydrochloric acid solution, with a concentration of 0.10 mol/L, assuming total acid ionization, is: Data (at 25 °C): [H⁺ ] [OH^- ] = 1,0 · 10^-14; pOH = -log[OH^- ]

a) 10^-13
b) 10^-1
c) 1
d) 7
e) 13

See Answer

Correct alternative: e) 13.

1st step: calculate the pH of the solution.

2nd step: convert to pOH value.

7. (Mackenzie) Water was added to 1.15 g of methanoic acid until 500 ml of solution was complete. Considering that at this concentration the degree of ionization of this acid is 2%, then the pOH of the solution is: Given the molar mass of methanoic acid = 46 g/mol

a) 2
b) 3
c) 12
d) 10
e) 11

See Answer

Correct alternative: e) 11.

1st step: calculate the molar concentration of acid.

2nd step: calculate the concentration of H ions⁺.

3rd step: calculate the pH of the solution.

4th step: transform the value to pOH.

pH and pOH in Enem

8. (Enem/2014) In order to minimize environmental impacts, Brazilian legislation determines that chemical residues released directly into the receiving body have a pH between 5.0 and 9.0. An aqueous liquid waste generated in an industrial process has a hydroxyl ion concentration equal to 1.0 x 10-10 mol/L. To comply with the legislation, a chemist separated the following substances, available in the company's warehouse: CH₃COOH, Na₂ONLY₄, CH₃OH, K₂CO₃ and NH₄Cl.

So that the residue can be released directly into the receiving body, which substance could be used to adjust the pH?

a) CH₃COOH
b) In₂ONLY₄
c) CH₃oh
d) K₂CO₃
e) NH₄Cl

See Answer

Correct alternative: d) K₂CO₃.

If the concentration of hydroxyl ions is equal to 1.0 x 10^-10 mol/L, the pOH of the residue is:

Consequently, the pH value is:

As the chemical residue must be released with a pH between 5 and 9, it is necessary that a substance with a basic character is added to raise the pH.

Analyzing the alternatives, we have to:

a) WRONG. Methanolic acid is a carboxylic acid and, therefore, has an acidic character. Its addition would further lower the pH.

b) WRONG. Sodium sulfate is a salt resulting from the reaction between strong acid and base.

Therefore, it is a neutral salt and its addition would not change the pH.

c) WRONG. Methanol is an alcohol and has a neutral character, so its addition would not change the pH.

d) CORRECT. Potassium carbonate is a salt resulting from the reaction between a weak acid and a strong base.

Its character is basic and is the most suitable for raising the pH of the residue by releasing OH ions^- in solution from the alkaline hydrolysis of the salt.

e) WRONG. Ammonium chloride is a salt resulting from the reaction between a strong acid and a weak base.

This salt has an acidic character and its addition would further lower the pH.

9. (Enem/2018) Red cabbage juice can be used as an acid-base indicator in different solutions. For this, just mix a little of this juice to the desired solution and compare the final color with the pH indicator scale, with values ​​from 1 to 14, shown below.

Using the acid-base indicator and the scale to determine the pH of human saliva and gastric juice, we have, respectively, the colors

a) red and red.
b) red and blue.
c) pink and purple.
d) purple and yellow.
e) purple and red.

See Answer

Correct alternative: e) purple and red.

A healthy mouth produces saliva with a pH around 7. It is a buffer solution of bicarbonate, biphosphate and monohydrogen phosphate so that the pH is practically constant.

Gastric juice, on the other hand, is composed of hydrochloric acid, a strong acid whose pH is close to 2.

Analyzing the alternatives, we have to:

a) WRONG. The red color indicates that both are acidic.

b) WRONG. This combination indicates that the substances are: acidic and basic.

c) WRONG. This combination indicates that the substances are: acidic and slightly basic.

d) WRONG. This combination indicates that the substances are: slightly basic and highly alkaline.

e) CORRECT. Saliva has a neutral pH and gastric juice has an acidic pH.

10. (Enem/2010) Decision to pave the MG-010 road, accompanying the introduction of exotic species, and the the practice of arson, threaten the sophisticated ecosystem of the rupestrian field of the Serra do reserve Ridge. Plants native to this region, highly adapted to a high concentration of aluminum, which inhibit root growth and hinder the absorption of nutrients and water, are being replaced by invasive species that would not naturally adapt to this environment, however they are dominating the edges of the highway, mistakenly called the "road ecological". Possibly the entry of exotic plant species in this environment was caused by the use, in this project, of a type of asphalt (cement-soil), which has a calcium-rich mixture, which caused chemical modifications to the soils adjacent to the highway MG-010.

_{Scientific American. Brazil. Year 7, nº 79. 2008 (adapted).}

This statement is based on the use of cement-soil, a mixture rich in calcium that

a) inhibits aluminum toxicity, raising the pH of these areas.
b) inhibits aluminum toxicity, reducing the pH of these areas.
c) increases aluminum toxicity, raising the pH of these areas.
d) increases aluminum toxicity, reducing the pH of these areas.
e) neutralizes aluminum toxicity, reducing the pH of these areas.

See Answer

Correct alternative: a) inhibits aluminum toxicity, raising the pH of these areas.

The calcium present in cement-soil is in the form of oxide, which in contact with water generates hydroxide.

Thus, the hydroxyl ions present in the medium raised the soil pH.

These species in contact with aluminum cause it to precipitate, decreasing its toxicity.

These chemical modifications in the soil enabled the growth of non-native plants in the region.

To acquire more knowledge, be sure to read these texts.:

• Chemical balance
• ionic balance
• Titration
• Exercises on chemical balance
• Exercises on inorganic functions