Question

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

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, $[\ce{Al^{3+}}]=2\times0.695\ \text{M}=1.39\ \text{M}$

Step3: Calculate $[\ce{SO4^{2-}}]$

From the equation, 1 mole of $\ce{Al2(SO4)3}$ gives 3 moles of $\ce{SO4^{2-}}$.
So, $[\ce{SO4^{2-}}]=3\times0.695\ \text{M}=2.085\ \text{M}$

Answer:

$[\ce{Al^{3+}}]=1.39\ \text{M}$; $[\ce{SO4^{2-}}]=2.085\ \text{M}$