Topic 4.3 - Acids And Bases 1

Question 1

Calculate the pH of a 0.0346 moldm^-3 solution of benzoic acid,

(K_A = 3.67 x 10^-4)

Answer 1

3.67 x 10^-4 x 0.0346 = [H₃O⁺]²

[H₃O⁺] = 3.56 x 10^-3

-log (3.56 x 10^-3) = 2.45

Question 2

Calculate the concentration of a solution of NaHSO₄ with pH 4.6 (K_A = 2.93 x 10^-6)

Answer 2

10^-4.6 = 2.51x10^-5

[H₃O+]² / K_A = [NaHSO₄]

(2.51 x 10^-5)² / 2.93 x 10^-6 = 2.15 x 10^-4 moldm^-3

Question 3

A 278cm³ solution of an acid HX contains 4.8g of the acid. If the pH is 2.4 and K_A is found to be 4.3 x 10^-4, calculate the molecular mass of the acid.

Answer 3

10^-2.4 = 3.98 x 10^-3

(3. 98 x 10^-3)² / 4.3 x 10^-4 = [HX]
(0. 0369 x 278) / 1000 = 0.0102
4. 8 / 0.0102 = 469

Question 4

Calculate the pH of a solution of 0.25moldm^-3 NaOH.

(K_W = 1 x 10^-14)

Answer 4

[NaOH] = [OH^-]

1 x 10^-14 / 0.25 = [H₃O⁺]

[H₃O⁺] = 4 x 10^-14

-log [4 x 10^-14] = 13.40

Question 5

Find the pH of a 3.0moldm^-3 solution of ethanoic acid.

(K_A = 2 x 10^-4)

Answer 5

[H₃O⁺]² = 2 x 10^-4 x 3

[H₃O⁺]² = 6 x 10^-4

[H₃O⁺] = 0.0245

• log [0.0245] = 1.61

Question 6

Work out the pH of a 10dm³ solution containing 3 moles of HCl.

Answer 6

3 / 10 = 0.3moldm^-3

[HCl] = [H₃O⁺]

-log [0.3] = 0.523

Question 7

A 2.52g sample of ethanoic acid was dissolved in water and made up to 250cm³. Calculate the pH of the acid given that K_A is 1.5x10^-5.

Answer 7

moles = 2.52 / 60

moles = 0.042

(0.042 x 1000) / 250 = conc

conc = 0.168

1.5 x 10^-5 x 0.168 = [H₃O⁺]²

[H₃O]⁺ = 0.00159

• log [0.00159] = 2.80

Question 8

What is the pH of a solution of H₃PO₄ when dissolved in water to form a solution of concentration 0.78moldm^-3?

Answer 8

3[H₃PO₄] = [H₃O⁺]

[H₃O⁺] = 2.34moldm^-3

• log 2.34 = -0.37

Question 9

Calculate the pH of propanoic acid given that the concentration is 0.02moldm^-3 and the dissociation constant for the acid is 6.7x10^-3.

Answer 9

[H₃O⁺]² = 6.7x10^-3 x 0.02

[H₃O⁺] = 0.0116

-log 0.016 = 1.94

Question 10

Given that the pH of a solution of sulphuric acid is -0.72, calculate the concentation of the acid.

Answer 10

10^0.72 = [H3O+] = 5.25

2[H₂SO₄] = [5.25]

[H₂SO₄] = 2.62moldm^-3

Question 11

Calculate a value for the dissociation constant of a sample of ethanoic acid with pH 6.08 and concentration 0.37moldm^-3.

Answer 11

10^-6.08 = [H₃O⁺] = 8.32 x 10^-7

[8.32 x 10^-7]² / 0.37 = 1.87 x 10^-12

Question 12

Considering the K_W of water is 8 x 10^-15 at 42^oC, calculate the pH of a 0.59moldm^-3 solution of Al(OH)_3.

Answer 12

[OH^-] = 3[Al(OH)₃] = 1.77

(8 x 10^-15) / 1.77 = [H₃O⁺] = 4.52 x 10^-15

• log (4.52 x 10^-15) = 14.34

Question 13

A 0.5dm³ solution contains 1.8g of weak acid and has a pH of 2.9. Calculate the M_R of the acid to 2 significant figures given that the K_A of the acid is 4.0 x 10^-5.

Answer 13

10^-2.9 = [H₃O⁺] = 0.00125

[0.00125]² / (4.0 x 10^-5) = 0.0396moldm^-3

0. 0396 x 0.5 = 0.0198mol
1. 8 / 0.0198 = 90.86

Rounded to 91

Question 14

Calculate the pH of boric acid (H₃BO₄)with concentration 0.76moldm^-3

Answer 14

[H₃O⁺] = 3[H₃BO₄] = 2.28

• log 2.28 = -0.358

Question 15

Calculate the pH of propanoic acid with concentration 0.34moldm^-3 and K_A = 7.75 x 10^-6

Answer 15

7.75 x 10^-6 x 0.34 = [H₃O⁺]²

[H₃O⁺] = 0.00162

• log 0.00162 = 2.79

Question 16

Define a Bronsted-Lowry acid.

Answer 16

A proton donor

Question 17

Define a Bronsted-Lowry base.

Answer 17

A proton acceptor

Question 18

Define a Lewis base.

Answer 18

A lone pair donor.

Question 19

Define a Lewis acid.

Answer 19

A lone pair acceptor.

Question 20

If a substance is described as amphoteric, what does it mean?

Answer 20

Something that can act as an acid or a base.

Question 21

A solution strong base has pH 10.3 at 17^oC. The solution was made from 2.4g of the base dissolved in 250cm³ of water. At 17^oC, K_W has a value of 6.41 x 10^-13. Calculate the molecular mass of the base to 3 significant figures.

Answer 21

10^-10.3 = 5.01 x 10^-11

(6.41 x 10^-13) / (5.01 x 10^-11) = [OH^-] = 0.0128

[OH^-] = [base]

(0. 0128 x 250) / 1000 = 0.00320
2. 4 / 0.00320 = M_R = 750.6

Rounded to 751