The pH scale and strong acids

Question 1

What do the numbers on the pH scale represent in terms of H+ concentration?

Answer 1

• pH 0 is 10^0 = 1.
• pH 1 is 10^-1.
• pH 15 is 10^-15.

Question 2

What is the relationship in terms of pH and the concentration of H+?

Answer 2

• A low value of H+ matches a high value of pH.
• A high value of H+ matches a low value of pH.

Question 3

What is the equation to link pH and H+?

Answer 3

1. pH = -log[H+ (aq)].
2. [H+ (aq)] = 10^-pH.

Question 4

As the pH number changes by 1, how much does the H+ concentration change?

Answer 4

By 10.

Question 5

What dilution would you need to dilute a solution of pH 1 to pH 4?

Answer 5

There are 3 pH units so would do 10 x 10 x 10 = 1000 times.

Question 6

What is the difference in H+ of pH 1 and pH 14?

Answer 6

10^13.

Question 7

How should you write pH values?

Answer 7

To 2 decimal places.

Question 8

How should you write [H+] concentration values?

Answer 8

To 3 significant figures.

Question 9

How do you represent how a strong/monobasic acid dissociates?

Answer 9

HA (aq) - H+ (aq) + A- (aq).

Question 10

As a strong acid completely dissociates in aqueous solutions, what is the relationship between [HA] and [H+]?

Answer 10

They are equal to one another.

Question 11

How can the pH of a strong acid be calculated?

Answer 11

Directly from the concentration of the acid.

Question 12

What is the process to calculate pH from acid concentration?
- ‘What is the pH of a sample of nitric acid that has a concentration of 1.35 x 10^-2 moldm-3’? (2)?

Answer 12

1. Convert [HA] into [H+]:
   - HNO3 is a strong acid so it completely dissociates.
   - [H+] = [HNO3] = 1.35 x 10^-2.
2. Find pH:
   - pH = -log[H+] = -log[1.35 x 10^-2] = 1.87.

Question 13

What is the process to calculate the concentration of strong acids from pH?
- ‘What is the concentration of HCl with a pH of 3.78’? (2)?

Answer 13

1. Find [H+]:
   - [H+] = 10^-pH = 10^-3.79 = 1.62 x 10^-4 moldm-3.
2. Convert [H+] into [HA]
   - HCl is a strong acid and completely dissociates.
   - [HCl] = [H+] = 1.62 x 10^-4 moldm-3.

Question 14

What does diluting a strong acid by half change the pH by?

Answer 14

Only 0.30.

Question 15

What is the process for working out the change in pH?
- ‘What is the change in pH if 50 cm3 of 0.100 moldm-3 HCl is diluted to 100 cm3 with water’? (2)?

Answer 15

1. Find the concentration of the diluted HCl:
   - On diluting to 100 cm3, the concentration has been halved to 0.0500 moldm-3.
2. Find the pH values before and after dilution:
   - HCl is a strong acid and completely dissociates:
   [H+] = [HCl].
   - Before dilution, [H+] = 0.100:
   pH = -log(0.100) = 1.00.
   - After dilution, [H+] = 0.0500
   pH = -log(0.0500) = 1.30.

Changing the concentration by a factor of 2 changes the pH by 0.30.

Question 16

Do the calculations differ between strong and weak acids?

Answer 16

Yes, these calculations above are all for strong acids, not weak acids.