You have the pKa values wrong. pK1 = 2.15; pK2 = 7.2 (which you have right) and pK3 = 12.1.
Use the Henderson-Hasselbalch equation.
pH = pk2 + log (base)/(acid)
7.5 = 7.2 + log (base)/(acid) or
7.5 = 7.2 + log b/a
H3PO4 + NaOH ==> NaH2PO4 + H2O
NaH2PO4 + NaOH ==> Na2HPO4 + H2O
Na2HPO4 + NaOH ==> Na3PO4 + H2O
The one you want is the middle one since that's the closest to your desired pH of 7.5. millimols H2PO4^- you have is 30 mL x 0.1M = 3.
.....H2PO4^- + OH^- ==> HPO4^2- + H2O
I.....3........0..........0........0
add............x...................
C.....-x......-x...........x
E.....3-x......0...........x
Substitute the E line into the HH equation and solve for x = millimols OH to be added to H2PO4^- to obtain the pH you want. Then convert that many millimols NaOH to mL using the M of the NaOH. This is not the end of the problem. Then you must add mL of 0.1M NaOH needed to neutralize the first H^+ from H3PO4. That is, it takes 30 mL of 0.1M NaOH to neutralize H3PO4 to H2PO4^- and that's where your starting with this problem so that 30 mL must added to whatever you calculate for the pH you want. Post your work if you get stuck. I don't think I've missed anything.
Calculate the volume of NaOH 0.1M that needs to be calculated to be added to a 30ml o-phosphoric acid 0.1 in order to prepare solutions of pH 7.5 and pH 6.9? pKa3= 2.15; pKa2=7.2; pKa1 = 12.1
2 answers
Thanks for d reply. Pls if i want to calculate for a pH less than 7 (eg 5.8), wil i subtract d volume i obtain from 30ml or will i add it?
1 answer