In a basic buffer, 0.0025 mole of NH_(4)Cl and 0.15 mole of NH_(4) OH are present. The pH of the solution will be (pK_(a)=4.74)

In a basic buffer, 0.0025 mole of NH_(4)Cl and 0.15 mole of NH_(4) OH

1.	In a basic buffer, 0.0025 mole of NH_(4)Cl and 0.15 mole of NH_(4) OH are present. The pH of the solution will be (pK_(a)=4.74)
Answer» 11.04 10.24 6.62 5.48 Solution :From Henderson's equation, `POH = pK_(b) + log. (["Salt"])/(["Base"])` `=4.74+log.(0.0025//V)/(0.15//V)=4.74+log. (1)/(60)` `=4.76-1.78=2.96`. HENCE, pH = 14 - pOH = 14 - 2.96 = 11.04.