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Date November 2019 Marks available 4 Reference code 19N.3.hl.TZ0.8
Level HL Paper 3 Time zone TZ0
Command term Apply and Determine Question number 8 Adapted from N/A

Question

1.40 × 10−3 g of NaOH (s) are dissolved in 250.0 cm3 of 1.00 × 10−11 mol dm−3 Pb(OH)2 (aq) solution.

Determine the change in lead ion concentration in the solution, using section 32 of the data booklet.

Markscheme

«[OH] = 1.40 × 10 3 g 40.00  g mo l 1 × 0.2500  d m 3 =» 1.40 × 10−4 «mol dm−3» ✔

«[OH] from dissolved Pb(OH)2 is negligible»

NOTE: Accept «ratio  [ P b 2 + ] i n i t i a l [ P b 2 + ] f i n a l  =» 13.7 OR  «ratio  [ P b 2 + ] f i n a l [ P b 2 + ] i n i t i a l  =» 0.0730 for M4.

 

Ksp = [Pb2+][OH]2
OR
1.43 × 10−20 = [Pb2+] × (1.40 × 10−4)2

[Pb2+]final = 7.30 × 10−13 «mol dm−3» ✔

NOTE: Award [4] for correct final answer.

 

«change in [Pb2+] = 1.00 × 10−11 − 7.30 × 10−13 =» 9.27 × 10−12 «mol dm−3» ✔

NOTE: Award [3] for correct [Pb2+]final.

Examiners report

[N/A]

Syllabus sections

Core » Topic 1: Stoichiometric relationships » 1.3 Reacting masses and volumes
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