Ksp

[destroythemcat]

The student determines the molar solubility, S, of PbI2(s) in water to be 1.89*10^-3.

Question: How many grams of lead(II)iodide are present in 100mL of a saturated aqueous solution?
A. 0.0410g
B. 0.0871g
C. 2.470g
D. 8.71g

The answer was B because 100mL=0.1L and 0.1L*1.89*10^-3*461g/mol=0.09g

Im confused....does Ksp=#moles of disolved ions/liter?? Can anyone help me out?

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[justadream]

@destroythemcat

You are given the MOLAR SOLUBILITY (not Ksp). I believe MS is the "x" that you solve for when you write the Ksp expression. Here, they already gave it to you.

I believe Molar solubility is like the # of moles that are soluble in 1 L of solution.

 

[destroythemcat]

Good catch! thank you!

 

[Teleologist]

Im confused....does Ksp=#moles of disolved ions/liter?? Can anyone help me out?

No that's not Ksp; that's a molarity value.