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Buffer index of a buffer of 0.1 M `NH_(4)OH` and 0.1 M `NH_(4) C l ` is `(pK_(b)` for `NH_(4)OH=4.74)`A. 0.116B. 0.232C. 0.058D. 0.348 |
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Answer» Correct Answer - A `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` |
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