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Help needed with Chemistry titration problem.?
19 2009:
100 mL of a soln that is 0.05M Sr^+2 and 0.05M Ra^+2 is titrated with 0.250 M SO4^-2. What is the pSO4 ^-2 after the addition of 48.6 mL of the base?
a. 2.65 b. 11.35 c. 10.39 d. 3.61
Explanation is necessary. Thank you.
1 Answer
- 1 decade agoFavorite Answer
you have added 0.250 M* 48.6/1000 L= 0.01215 Moles of SO4 --
sulphates become insoluble as you go down group 2 so all the Ra should be turned into RaSO4
that is = 0.05 * 100/1000 = 0.005 Moles
so that gets you down to 0.01215 - 0.005 moles of SO4-- left = 0.00715Moles
The Sr will also for SrSO4 so use up more SO4--
Amount of SrSO4 = 0.05 * 100/1000 = 0.005 moles
so the amount of SO4-- left is 0.00715 - 0.005 = 0.00215Moles of SO4 --
Now I am confused
Are you asking for the percentage SO4--?
ans which is the Base?? what Base
I thought I had understood what you were after but it looks like I didn't
Anyway hope this helps you understand what you need to do to finish it off
S
PS Let me know how you get on:)