See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

(A) 3 2 3 4 4 2 4 4 H PO PO H PO HPO Initial m. mol 10 10 0 0 m.mol. after neutralization 0 0 10 10       So,   2 a pH pK 8 Buffer   (B)             3 3 4 2 4 4 H PO H PO PO 10 10 10   3 2 3 4 4 2 4 4 H PO PO H PO HPO Initial m. mol 10 10 10 0 After neutralization 0 0 20 10       Again it is a buffer 2 4 2 2 2 a HPO 10 pH pK log 8 log