DP Chemistry (last assessment 2024)

Test builder »

Question 19N.3.hl.TZ0.8

Select a Test
Date November 2019 Marks available [Maximum mark: 4] Reference code 19N.3.hl.TZ0.8
Level hl Paper 3 Time zone TZ0
Command term Apply, Determine Question number 8 Adapted from N/A
8.
[Maximum mark: 4]
19N.3.hl.TZ0.8

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.

[4]

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.

Syllabus sections