Turn on thread page Beta

how to do this chemistry question? EDEXCEL AS watch

    • Thread Starter

    Name:  Capture.JPG
Views: 19
Size:  54.9 KB

    I don't even know where to start off for both a) and b)

    • Thread Starter


    i got a) but dunno ow to do b)

    On addition of iodide, the liberated copper(ii) cations react to form iodine (Eq. 1) which becomes the analyte of the titration.
    A mean titre of 20.50 cm3 is equivalent to 2.255e-3 mol of thiosulfate added.
    From the stoichiometry ratio (Eq. 2), the moles of iodine are half those of thiosulfate (1.1275e-3 mol).

    Back to Eq. 1, the amount of copper(ii) cations is twice the amount of iodine (2.255e-3 mol).

    Hydrated copper(ii) sulfate dissolves in water like this:
    CuSO4.nH2O(s) + aq → Cu2+(aq) + SO42-(aq) + nH2O(l)
    The amount of crystal is equal to the amount of copper(ii).
    However, since just 1/10 of the solution was used, the total amount of crystal is 2.255e-2 mol.

    2.255e-2 mol of CuSO4 has mass = 3.60 g. The rest out of 5.63 g must therefore be water (2.03 g = 0.11 mol).
    If 2.255e-2 mol carries 0.11 mol of water, n = 5. The crystal is pentahydrate.
Submit reply
Turn on thread page Beta
Updated: March 1, 2018

University open days

  • Heriot-Watt University
    School of Textiles and Design Undergraduate
    Fri, 16 Nov '18
  • University of Roehampton
    All departments Undergraduate
    Sat, 17 Nov '18
  • Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 17 Nov '18
Have you ever experienced bullying?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.