We need to make 350mL of a solution with 3mM Lead Nitrate, 50mM of Tris Base, and 0.01M KH2PO4. How much of each component are needed (including water)?

Question

We need to make 350mL of a solution with 3mM Lead Nitrate, 50mM of Tris Base, and 0.01M KH2PO4. How much of each component are needed (including water)?

Lead (II) Nitrate MW: 331.2 g/mole

Tris Base MW: 121.14 g/mole

KH2PO4 0.1M Solution

Answer 1

Hi Erika N.!

The mass required for the first two items listed can be obtained by simple dimensional analysis:

3mmol Pb(NO₃)₂*(1 mol/1000mmol)*(331.2 g Pb(NO₃)₂/1 mol Pb(NO₃)₂)

mass = 0.9936g Pb(NO₃)₂

50mmol Tris*(1 mol/1000mmol)*(121.14g Tris/1 mol Tris)

mass = 6.057g Tris

Alternatively, you may omit the conversion from millimole into moles altogether if you wish. For example:

3mmol Pb(NO₃)₂*(0.3312 g Pb(NO₃)₂/1 mmol Pb(NO₃)₂)

mass = 0.9936g Pb(NO₃)₂

Notice that this solution yields the same answer.

Because the KH₂PO₄ is a stock solution preparation, one way to find the volume required is by using the dilution equation:

M₁V₁ = M₂V₂ => V₁ = M₂V₂/M₁

<=> (0.01M KH₂PO₄*350mL)/0.1M KH₂PO₄

V₁ = 35mL of 0.1M KH₂PO₄

This volume represents the volume of stock solution we need for dilution into our buffer.

Additionally, you may have noticed your stock solution and buffer solution differ only by 1 order of magnitude in terms of KH₂PO₄, or by a multiplication or division of 10. Another way to say this is that the buffered solution represents a 10X (10-fold) dilution of your stock solution. Typically, this is done intentionally beforehand to make calculation convenient. Rather than using the dilution equation, we could instead simply divide the volume of the buffered solution by 10, and we then obtain the volume of stock 0.1M KH₂PO₄ solution required for this buffer.

Hope this helps!

Cheers