DP Chemistry (last assessment 2024)

Test builder »

Question 19M.3.sl.TZ2.16

Select a Test
Date May 2019 Marks available [Maximum mark: 4] Reference code 19M.3.sl.TZ2.16
Level sl Paper 3 Time zone TZ2
Command term Determine, Outline Question number 16 Adapted from N/A
16.
[Maximum mark: 4]
19M.3.sl.TZ2.16

Excess acid in the stomach can cause breakdown of the stomach lining.

(a(i))

Outline how ranitidine (Zantac) inhibits stomach acid production.

[1]

Markscheme

blocks/binds H2/histamine receptors «in cells of stomach lining»
OR
prevents histamine molecules binding to H2/histamine receptors «and triggering acid secretion»  [✔]

Examiners report

This question was not very well answered even though it has been seen in previous scripts. Some students were not through enough in their answers to receive a mark while others confused the action of ranitidine which blocks H2 receptors with omeprazole which is a proton pump inhibitor.

(a(ii))

Outline two advantages of taking ranitidine instead of an antacid which neutralizes excess acid.

[2]

Markscheme

Any two of:
ranitidine can be effective in treating ulcers «but antacid is not»  [✔]

ranitidine can prevent long-term damage «from overproduction of acid and antacid does not»  [✔]

ranitidine has a long-term effect «and antacid has short-term effect only»  [✔]

ranitidine does not affect ionic balance in body «and antacid does» [✔]

ranitidine does not produce bloating/flatulence [✔]

 

Note: Accept “ranitidine stops the over production of acid in the stomach while antacids neutralize the excess acid giving temporary relief”.

Examiners report

While most students were awarded at least one of the two marks possible for this question some of the descriptions were too vague or incomplete to receive a mark.

(b)

Some antacids contain carbonates.

Determine the pH of a buffer solution which contains 0.160 mol dm−3 CO32− and 0.200 mol dm−3 HCO3−, using section 1 of the data booklet.

pKa (HCO3−) = 10.32

[1]

Markscheme

«pH = pKa + log  [ A − ] [ HA] = 10.32 + log  0.160 0.200 = 10.32 – 0.097»

«pH =»10.22  [✔]

Examiners report

This was in general a well answered question. Most candidates who did not receive the mark inverted the concentration of the conjugate base/concentration of the acid in the calculation.

Syllabus sections