what is the concentration of ions in a solution that is 0.695 m in aluminum sulfate, al₂(so₄)₃? al³⁺ =…
what is the concentration of ions in a solution that is 0.695 m in aluminum sulfate, al₂(so₄)₃? al³⁺ = select m so₄²⁻ = select m
Answer
Explanation:
Step1: Write dissociation equation
$$\ce{Al2(SO4)3 -> 2Al^{3+} + 3SO4^{2-}}$$
Step2: Calculate $[\ce{Al^{3+}}]$
From the equation, 1 mole of $\ce{Al2(SO4)3}$ gives 2 moles of $\ce{Al^{3+}}$. So, $[\ce{Al^{3+}}]=2\times0.695\ M = 1.39\ M$
Step3: Calculate $[\ce{SO4^{2-}}]$
From the equation, 1 mole of $\ce{Al2(SO4)3}$ gives 3 moles of $\ce{SO4^{2-}}$. So, $[\ce{SO4^{2-}}]=3\times0.695\ M = 2.085\ M$
Answer:
$[\ce{Al^{3+}}]=1.39\ M$ $[\ce{SO4^{2-}}]=2.085\ M$