How many moles of NaF must be dissolved in 1.00 liter of a saturated solution of PbF 2 at 25°C to reduce the [Pb 2+] to 1.0 × 10 –6 M? The K sp for PbF 2 at 25 °C is 4.0 × 10 –8.

0.1957 moles of NaF

Explanation:

The Pb²⁺ and F⁻ are in equilibrium with PbF₂ as follows:

PbF₂(s) ⇄ Pb²⁺(aq) + 2F⁻(aq)

Where Ksp expression is:

Ksp = 4.0x10⁻⁸ = [Pb²⁺] [F⁻]²

A saturated solution contains the maximum possible amount of Pb²⁺ and F⁻. That is:

PbF₂(s) ⇄ Pb²⁺(aq) + 2F⁻(aq)

PbF₂(s) ⇄ X + 2X

Where X is amount of ions presents in solution

4.0x10⁻⁸ = [Pb²⁺] [F⁻]²

4.0x10⁻⁸ = [X] [2X]²

4.0x10⁻⁸ = 4X³

4.0x10⁻⁸/4 = X³

1.0x10⁻⁸ = X³

2.15x10⁻³M = X

That means initial concentration of Pb²⁺ is = X = 2.15x10⁻³M and [F⁻] = 2X = 4.30x10⁻³M

Now, using again Ksp, if you want a [Pb²⁺] = 1.0x10⁻⁶M, the [F⁻] you need is:

4.0x10⁻⁸ = [Pb²⁺] [F⁻]²

4.0x10⁻⁸ = [1.0x10⁻⁶M] [F⁻]²

0.04M = [F⁻]²

0.2M = [F⁻]

You need a final concentration of 0.2M of F⁻. As initial concentration was 4.30x10⁻³M and volume of the buffer is 1.00L, the moles of F⁻ = moles of NaF you must add are:

0.2M - 4.30x10⁻³M =

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