First of all - you have already wrote
Henderson-Hasselbalch equation:
pH=pK
a+log([A
-]/[HA])
Note, that when [A
-]=[HA] pH=pK
a. So if you know that H
2PO
4-/HPO
42- is used for buffers with pH close to 7, you have to look through the answers.
3.10ml 1 M H3PO4 + 5ml 1 M NaOH
2.10ml 1 M H3PO4 + 10 ml 1 M NaOH
1.10ml 1 M H3PO4 + 15 ml 1 M NaOH
4.10ml 1 M H3PO4 + 25ml 1 M BAOH
(I have sorted them slightly different then in original question).
Solution with 5mL of base contains equimolar buffer made of H
3PO
4 and H
2PO
4-.
Solution with 10mL of base contains only H
2PO
4- - no buffer here.
Solution with 15mL of base contains equimolar buffer made of H
2PO
4- and HPO
42-.
Solution with 25mL of base contains equimolar buffer made of HPO
42- and PO
43-.
Of those four solutions only one can have pH close to 7.