Question

Given 1L of 0.1M stock solutions of NaH2PO4 and Na2HPO4, and ample water, describe in as much detail as possible how you would go about making a 0.05M solution of phosphate buffer with a pH=6.75?

Answer 1

Given 1L of 0.1M stock solutions of NaH2PO4 and Na2HPO4, and ample water, describe in as much detail as possible how you would go about making a 0.05M solution of phosphate buffer with a pH=6.75

Here in this question Na2HPO4 will behave as base and NaH2PO4 will behave as acid.

Na2HPO4àà 2Na⁺ + HPO4^2-

NaH2PO4àà Na⁺ + H2PO4^-

And then; according to Bronsted-Lowry's theory:

H2PO4^- àà H⁺ + HPO4^2-

Complete equation can be written as Na2HPO4 àà Na⁺ + H⁺ + HPO4^2-

We can use handerson hasselbalch equation,

pH = pKa +log [Na2HPO4]/[NaH2PO4]

pKa of phosphate buffer is 7.21

6.75 = 7.21 + log [Na2HPO4]/[NaH2PO4]

log [Na2HPO4]/[NaH2PO4] = -0.46

[Na2HPO4]/[NaH2PO4] = -antilog 0.46

[Na2HPO4]/[NaH2PO4] =0.3467

We need [Na2HPO4] + [NaH2PO4] = 0.05

By solving above two equations we get,

[Na2HPO4]= 0.3467[NaH2PO4]

0.3467[NaH2PO4] + [NaH2PO4] = 0.05

[NaH2PO4] = 0.05/1.3464 = 0.037

[Na2HPO4] = 0.05-0.037 = 0.012

This means you need 0.037 moles of NaH2PO4 and 0.012moles of Na2HPO4 in 1lit of solution

Now moles= molarity *volume

Therefore volume= 0.037/0.1=0.37 liter of NaH2PO4

And = 0.012/0.1= 0.12liter of Na2HPO4

370ml of NaH2PO4 and 120liter of Na2HPO4 should be mix in 1 liter of water to get desired concentration of buffer with pH 6.75

But molarity of the solution is given as 0.1M