Write up the recipe to make 100mL of a 200mM phosphate buffer, pH 7.0 by using:

(a) only NaH2PO4 (MW 120g/mol) and NaOH (1.00M)

(b) only Na2HPO4 (MW 142g/mol) and HCl (1.00M)

(c) only Na2HPO4 and NaH2PO4

Please help me. I don't seem to understand how to even start......

Thanks

I didn't answer this last yesterday when you first posted because the post was screwed up. It looks ok now.

You want 100 mL of 200 mM phosphate buffer. That is 100 mL x 0.2 M = 20 millimols phosphate. So you know base(b) + acid (a) = 20. That is equation 1.

You obtain equation 2 from the HH equation of 7.00 = pKa2 + log (b/a) and solve that equation for b/a. That is equation 2. pKa2 is that for H3PO4.

Solve equation 1 and equation 2 simultaneously and determine b and a (they will be in millimols).

For c, which is the easiest of the 3, now that you know b and a (and those are in millimoles), calculate how much NaH2PO4 and Na2HPO4 that you must weigh out for the millimoles of each that you calculate. Obviously NaH2PO4 is the acid and Na2HPO4 is the base.

Then for part a and part b you go a slightly different route. I'll start you on part a and part b I'll leave to you. It's the same kind of thing.

......H2PO4^- + OH^- ==> HPO4^2- + H2O
I......20.......0..........0
add.............x............
C.....-x.......-x...........x
E......20-x.....0...........x
You are starting with 20 millimoles of the acid and x = millimols NaOH to be added.

Then substitute the E line into the HH equation like this
7.0 = pKa2 + log (x)/(20-x)
Solve for x = millimols NaOH. Now you know mmols NaOH and you know it is 1 M, solve for mL NaOH to add to that 20 mmols of the NaH2PO4.

For part B, you start with 20 millimols Na2HPO4 and add H^+ to make H2PO4^- but it's done exactly the same way but now you let x = millimols HCl that must be added.
I love these problems. By the way, the CORRECT way to do these is with concentrations but I've usee millimols. You can use concentration, mols, millimols BECAUSE (a) it is a ratio and (b) M = millimols/mL and since mL is the same for both acid and base and since they are in the denominator of base/acid, that means the denominator ALWAYS cancels so I just don't put the mL there in the first place.

Thank you so much!

Sure! Let's go through each scenario step by step.

(a) To make a phosphate buffer using only NaH2PO4 and NaOH, you need to calculate the amount of NaH2PO4 and NaOH needed to achieve the desired pH and concentration.

First, let's calculate the moles of NaH2PO4 needed:
moles of NaH2PO4 = concentration (in M) x volume (in L)
= (200 mM / 1000 mM per M) x (0.100 L)
= 0.020 moles

Next, let's calculate the moles of NaOH needed to adjust the pH:
Since NaH2PO4 is a monobasic acid, we can assume that it will react with an equal amount of NaOH to form the corresponding salt, Na2HPO4.

moles of NaOH = moles of NaH2PO4
= 0.020 moles

To calculate the volume of NaOH solution required, we use the formula:
volume (in L) = moles / concentration (in M)

volume of NaOH = 0.020 moles / 1.00 M
= 0.020 L or 20 mL

So, to make 100 mL of a 200 mM phosphate buffer at pH 7.0 using only NaH2PO4 and NaOH, you would need to dissolve 0.020 moles of NaH2PO4 in 80 mL of water, and then add 20 mL of 1.00 M NaOH to adjust the pH to 7.0.

(b) To make a phosphate buffer using only Na2HPO4 and HCl, the process is similar.

First, calculate the moles of Na2HPO4 needed:
moles of Na2HPO4 = concentration (in M) x volume (in L)
= (200 mM / 1000 mM per M) x (0.100 L)
= 0.020 moles

Since Na2HPO4 is a dibasic salt, and HCl is a monobasic acid, we can assume that each mole of Na2HPO4 will react with two moles of HCl to form the corresponding salt, NaH2PO4.

moles of HCl = 2 x moles of Na2HPO4
= 2 x 0.020 moles
= 0.040 moles

To calculate the volume of HCl solution required, we use the same formula as before:
volume (in L) = moles / concentration (in M)

volume of HCl = 0.040 moles / 1.00 M
= 0.040 L or 40 mL

So, to make 100 mL of a 200 mM phosphate buffer at pH 7.0 using only Na2HPO4 and HCl, you would need to dissolve 0.020 moles of Na2HPO4 in 60 mL of water, and then add 40 mL of 1.00 M HCl to adjust the pH to 7.0.

(c) To make a phosphate buffer using only Na2HPO4 and NaH2PO4, the process is slightly different.

First, calculate the ratio of Na2HPO4 to NaH2PO4 needed to achieve the desired pH.

pKa of the phosphate buffer system is around 7.2, which means at pH 7.0, the buffer exists as a mixture of Na2HPO4 and NaH2PO4 in approximately equal amounts.

So the ratio of Na2HPO4 to NaH2PO4 is approximately 1:1.

Next, calculate the moles of Na2HPO4 and NaH2PO4 needed:
moles of Na2HPO4 = concentration (in M) x volume (in L)
= (200 mM / 1000 mM per M) x (0.100 L)
= 0.020 moles

moles of NaH2PO4 = moles of Na2HPO4
= 0.020 moles

To calculate the volume of Na2HPO4 and NaH2PO4 solutions required, we use the same formula as before:
volume (in L) = moles / concentration (in M)

volume of Na2HPO4 = 0.020 moles / concentration (in M)
(concentration is not given in this case)

volume of NaH2PO4 = 0.020 moles / concentration (in M)
(concentration is not given in this case)

So, to make 100 mL of a 200 mM phosphate buffer at pH 7.0 using only Na2HPO4 and NaH2PO4, you would need to dissolve 0.020 moles of Na2HPO4 and 0.020 moles of NaH2PO4 in water. The specific volumes of each solution will depend on the concentration (in M) of the stock solutions provided.

I hope this clarifies the process for you!