What Is the Concentration of SO4²⁻ When Al³⁺ Is 0.28M?

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SUMMARY

The concentration of sulfate ions (SO4²⁻) in an aqueous solution of aluminum sulfate (Al2(SO4)3) can be determined from the concentration of aluminum ions (Al³⁺). Given that the concentration of Al³⁺ is 0.28M, the stoichiometry of the dissociation reaction indicates that 1 mole of Al2(SO4)3 produces 3 moles of SO4²⁻. Therefore, the concentration of SO4²⁻ is calculated as 0.28M Al³⁺ multiplied by 3, resulting in a concentration of 0.84M SO4²⁻.

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Homework Statement


The concentration of Al3+ ion in aqueous solution of Al2(SO4)3 is 0.28M, Then the concentration of SO42- ion in this solution will be?


Homework Equations


Al2(SO4)3 → 2Al3+ + 3SO42-


The Attempt at a Solution


What should i compare to start of with?
 
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How many moles of sulphate ions are present in 1 mole Aluminium Sulphate?
 
3 moles, is this the answer?
AGNuke said:
How many moles of sulphate ions are present in 1 mole Aluminium Sulphate?
 
xiphoid said:
3 moles, is this the answer?

How many moles of Aluminium Sulphate do you currently have?
 
1 mole
Pranav-Arora said:
How many moles of Aluminium Sulphate do you currently have?
 
Take a look at the formula - how many moles of Al3+ and how many moles of SO42- in 1 mole of Al2(SO4)2? What is their ratio? Can you use this ratio to solve the problem?
 

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