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

🤓 Based on our data, we think this question is relevant for Professor Bindell's class at UCF.

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:


0.0490 g AgIO31 L×1 mol AgIO3282.80 g AgIO31.73 × 10–4 mol/L

View Complete Written Solution
Problem Details

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

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Ksp concept. You can view video lessons to learn Ksp. Or if you need more Ksp practice, you can also practice Ksp practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Bindell's class at UCF.