Question

solid al(no₃)₃ is added to distilled water to produce a solution in which the concentration of nitrate, no₃⁻, is 0.10 m. what is the concentration of aluminum ion, al³⁺, in this solution?
a 0.010 m
b 0.033 m
c 0.10 m
d 0.30 m

Explanation:

Step1: Write the dissociation equation

When \(Al(NO_3)_3\) dissociates in water, the equation is \(Al(NO_3)_3(s)
ightarrow Al^{3 +}(aq)+3NO_3^-(aq)\).

Step2: Analyze the ratio of ions

From the dissociation equation, the mole ratio of \(Al^{3+}\) to \(NO_3^-\) is \(1:3\). Let the concentration of \(Al^{3+}\) be \(x\) and the concentration of \(NO_3^-\) be \(y\). Then \(y = 3x\).

Step3: Calculate the concentration of \(Al^{3+}\)

Given \(y=[NO_3^-]=0.10M\). Substitute \(y = 3x\) into it, we get \(x=\frac{[NO_3^-]}{3}\).

$$x=\frac{0.10M}{3}\approx0.033M$$

Answer:

B. \(0.033M\)