Buffer index of a buffer of 0.1 M NH_(4)OH and 0.1 M NH_(4) C lis (pK_(b) for NH_(4)OH=4.74)

Buffer index of a buffer of 0.1 M NH_(4)OH and 0.1 M NH_(4) C lis (pK_

1.	Buffer index of a buffer of 0.1 M NH_(4)OH and 0.1 M NH_(4) C lis (pK_(b) for NH_(4)OH=4.74)
Answer» 0.116 0.232 0.058 0.348 Solution :`pOH = pK_(b) + log. (["SALT"])/(["Base"])=4.74` Suppose 1 ML of 1 M HCl (i.e. 0.001 MOL of HCl) is added. It will convert 0.001 mol of `NH_(4)OH` into `NH_(4)Cl` Now `[NH_(4)Cl]=0.1+0.001=0.101 M ` and `[NH_(4)OH]=0.1-0.001 = 0.099 M` Now, `pOH = 4.74 + log .(0.101)/(0.099)= 4.74+ 0.0086` Thus, change in pH = 0.0086 `:.` Buffer index `=(dn)/(dpH)=(0.001)/(0.0086) = 0.116`