See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

(P) Mg(OH)2(s)  Mg2+ + 2OH s 2s Ksp = 4s3 = 4  10-12 s = 10-4 [OH] = 2  10-4  pH = 10 + log 2 (Q) Basic buffer is formed. Hence, pOH = 1 log 2 b pk  = 4 – log 2 pH = 10 + log 2 (R) A salt of weak acid and weak base is formed in which ka = kb hence, pH = 7 or pH = 1 1 1 2 2 2 w a b pk pk pk   (S) H3PO4  H+ + 2 4 H PO 2 4 4 H PO H HPO      4 4 HPO H PO      pH of 0.01 M Na2HPO4 = 2 3 2 a a pk pk  = 8.102 12.5 2  pH = 20.602