The solubility of Ag2NtO4, silver nortonate, in pure water is 4.0 x 10-5 moles per liter. Calculate the value of Ksp for silver nortonate.
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To calculate the value of Ksp for silver nortonate (Ag2NtO4), you first need to write the balanced equation for the dissolution of silver nortonate in water.
The balanced equation for the dissolution of Ag2NtO4 in water can be represented as:
2Ag2NtO4(s) ⟶ 4Ag+(aq) + 2NtO4-(aq)
From this balanced equation, you can see that one mole of Ag2NtO4 produces 4 moles of Ag+(aq) ions and 2 moles of NtO4-(aq) ions.
The solubility of Ag2NtO4 in pure water is given as 4.0 x 10-5 moles per liter. This means that for every liter of water, 4.0 x 10-5 moles of Ag2NtO4 can dissolve.
Now, you can set up the equilibrium expression for the dissolution of Ag2NtO4 and express it in terms of its solubility (x):
Ksp = [Ag+]^4 [NtO4-]^2
Since one mole of Ag2NtO4 produces 4 moles of Ag+(aq) ions in the reaction, the concentration of Ag+ ions is 4 times the solubility of Ag2NtO4 in pure water:
[Ag+] = 4x
Similarly, the concentration of NtO4- ions is 2 times the solubility of Ag2NtO4:
[NtO4-] = 2x
Plugging these values in the equilibrium expression:
Ksp = (4x)^4 (2x)^2
Now, substitute the given solubility value (4.0 x 10-5 moles per liter) into the equation:
Ksp = (4 * 4.0 x 10-5)^4 (2 * 4.0 x 10-5)^2
Simplifying this expression will give you the value of Ksp for silver nortonate.