Question

Determine how you would make 3 mls of each buffer (12, 25, 75, and 150 mM NaCl) using 2 mM and 300 mM NaCl stock solutions.

Answer 1

There's something you should know, buffer of NaCl are rarely used because it does not have enough acid or base strength cause it's a salt. So is the real question a buffer of NaCl or maybe just prepare solutions on this?.

Either way, For all the buffers, do you actually need to use both solutions on every buffer? that's my doubt to continue the problem.

In general terms you usually use the general equation of HH to get the buffer solution:

pH = pKa + log [S]/[A]

But we don't have either the pH or pKa, so we can't solve it with this.

So, I'm gonna solve it with the way I know, of preparing solutions of salts.

we need a 3 mL of 12, 25, 75 and 150 mM of NaCl using 2 mM and 300 mM

moles of each solution:

moles 1 = 0.003 L * 0.012 mol/L = 3.6x10^-5 moles

moles 2 = 0.003 L * 0.025 mol/L = 7.5x10^-5 moles

moles 3 = 0.003 L * 0.075 mol/L = 2.25x10^-4 moles

moles 4 = 0.003 L * 0.150 mol/L = 4.5x10^-4 moles

With this, we can calculate the volume needed of the stock solution to prepare this solutions. If we take and assume that concentrations are additive then, 0.3 + 0.002 = 0.302 M so

V1 = 3.6x10^-5 moles / 0.302 moles/L = 1.192x10^-4 L or simply 0.119 mL

V2 =  7.5x10^-5 moles / 0.302 moles/L = 2.483x10^-4 L or simply 0.248 mL

V3 =  2.25x10^-4 moles / 0.302 moles/L = 7.450x10^-4 L or simply 0.745 mL

V4 =  4.50x10^-4 moles / 0.302 moles/L = 1.49x10^-3 L or simply 1.49 mL

Hope this helps