Notes: Ag2SO4 has a solubility of 4,496 g/L. What's its Ksp (product of solubility)?

Thursday, June 25, 2015

Ag2SO4 has a solubility of 4,496 g/L. What's its Ksp (product of solubility)?

Silver sulfate has the formula $\displaystyle{A}{g}_{{2}}{S}{O}_{{4}}$ . The number you gave for solubility is not within the range of solubility of silver sulfate, so I'm assuming that the comma should be a decimal point and you're trying to calculate Ksp based on a solubility of 4.496 grams per liter.

First, we need to change the grams to moles to get the molar solubility of $\displaystyle{A}{g}_{{2}}{S}{O}_{{4}}$ :

(4.496 grams Ag2SO4)/1 Liter x (1 mole/311.8 grams) = 0.01442 moles/L

Here's the equation for the dissociation of $\displaystyle{A}{g}_{{2}}{S}{O}_{{4}}$ :

$\displaystyle{A}{g}_{{2}}{S}{O}_{{4}}\to{2}{A}{{g}}^{+}+{S}{{O}_{{4}}^{{{2}-}}}$

The expression for the solubility product constant is:

$\displaystyle{K}_{{{s}{p}}}={{\left[{A}{{g}}^{+}\right]}}^{{2}}{\left[{S}{{O}_{{4}}^{{{2}-}}}\right]}$

For each mole that dissolves, two moles of $\displaystyle{A}{{g}}^{+}$ and 1 mole of $\displaystyle{S}{{O}_{{4}}^{{{2}-}}}$ are produced. We can substitute the molarity previously calculated, giving us:

$\displaystyle{K}_{{{s}{p}}}={\left[{2}{{\left({0.01442}\right]}}^{{2}}{\left[{0.01442}\right]}={1.199}{X}{{10}}^{{-{5}}}\right.}$

This is the accepted value of Ksp for silver sulfate at 25ºC as per the attached link, so this verifies that the solubility in the problem is 4.496 g/L and not 4,496 g/L.

Notes

Thursday, June 25, 2015

Ag2SO4 has a solubility of 4,496 g/L. What's its Ksp (product of solubility)?

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