# Problem: If 0.0490 g of AgIO3 dissolves per liter of solution, calculate the solubility-product constant.

###### FREE Expert Solution

The dissociation of AgIO3 in water is as follows:

AgIO3(s)  Ag+(aq) + IO3(aq)

The molar mass of AgIO3 is:

AgIO3        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:

1.73 × 10–4 mol/L

83% (437 ratings) ###### Problem Details

If 0.0490 g of AgIO3 dissolves per liter of solution, calculate the solubility-product constant.

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