In a basic buffer 0.0025 mole of NH4Cl and 0.15 mole of NH4OH are present. The pH of the solution will be[pKa= 4.74] 1) 11.042) 10.24 3) 6.624) 5.48

Taking pKb=4.74 [a misprint in the given data]Using formula- p(OH) = pKb + log(salt/base) = 4.74 + log(0.0025/0.15) = 4.74 + log(0.01666) = 4.74 + [-1.78] = 4.74 - 1.78 = 2.96So, p(OH) = 2.96Since, pH + p(OH) = 14pH = 14 - p(OH) = 14 - 2.96 = 11.04So, the pH of the given solution comes out to be 11.04