The dissociation of AgIO_{3} in water is as follows:

AgIO_{3}_{(s)} ⇌ Ag^{+}_{(aq)} + IO_{3}^{–}_{(aq)}

The molar mass of AgIO_{3} is:

AgIO_{3}**1** Ag × 107.9 g/mol Ag = 107.90 g/mol

**1** I × 126.9 g/mol I = 126.90 g/mol

**3** O × 16.00 g/mol O = __ 48.00 g/mol __

**Total**:** ****282.80 g/mol**

Converting **g/L to mol/L**:

$\frac{\mathbf{0}\mathbf{.}\mathbf{0490}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{AgIO}}_{\mathbf{3}}}}{\mathbf{1}\mathbf{}\mathbf{L}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{AgIO}}_{\mathbf{3}}}{\mathbf{282}\mathbf{.}\mathbf{80}{\mathbf{}}\overline{)\mathbf{g}\mathbf{}{\mathbf{AgIO}}_{\mathbf{3}}}}$**= ****1.73 × 10 ^{–4} mol/L**

If 0.0490 g of AgIO_{3} dissolves per liter of solution, calculate the solubility-product constant.

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