Respuesta :
Answer: The concentration of sulfate ions in the solution is 0.0522 M
Explanation:
To calculate the the molarity of solution:, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
Given mass of alum = 2.17 g
Molar mass of alum = 478.39 g/mol
Volume of solution = 175 mL
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{2.17\times 1000}{474.39\times 175}\\\\\text{Molarity of solution}=0.0261M[/tex]
The chemical equation for the ionization of alum follows:
[tex]KAl(SO_4)_2.12H_2O\rightarrow K^++Al^{3+}+2SO_4^{2-}+12H_2O[/tex]
1 mole of alum produces 1 mole of potassium ions, 1 mole of aluminium ions, 2 moles of sulfate ions and 12 moles of water
So, concentration of sulfate ions = [tex](2\times 0.0261)=0.0522M[/tex]
Hence, the concentration of sulfate ions in the solution is 0.0522 M
