Question

The molar solubility of AgBr in pure water is 7.3x10^-7 M. Calculate the Ksp.

The molar solubility of AgBr in pure water is 7.3x10^-7 M. Calculate the Ksp.

Homework Answers

Answer #1

Determine the Ksp of silver bromide, given that its molar solubility is 7.3 x 10¯7 moles per liter.

When AgBr dissolves, it dissociates like this:

AgBr (s) ⇌ Ag+ (aq) + Br¯ (aq)

The Ksp expression is:

Ksp = [Ag+] [Br¯]

There is a 1:1 molar ratio between the AgBr that dissolves and Ag+ that is in solution. In like manner, there is a 1:1 molar ratio between dissolved AgBr and Br¯ in solution. This means that, when 7.3 x 10¯7mole per liter of AgBr dissolves, it produces 7.3 x 10¯7 mole per liter of Ag+ and 7.3 x 10¯7 mole per liter of Br¯ in solution.

Putting the values into the Ksp expression, we obtain:

Ksp = (7.3 x 10¯7) (7.3 x 10¯7) = 5.329 x 10¯13

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