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Question: Buffer index of a buffer of 0.1 M \(N{{H}_{4}}OH\) and 0.1 M \(N{{H}_{4}}Cl\) is: A. 0.052 B. 0....

Buffer index of a buffer of 0.1 M NH4OHN{{H}_{4}}OH and 0.1 M NH4ClN{{H}_{4}}Cl is:
A. 0.052
B. 0.115
C. 0.025
D. 0.230

Explanation

Solution

Buffer Solution is defined as a water solvent based solution which consists of a mixture containing a weak acid and the conjugate base of the weak acid or a weak base and the conjugate acid of the weak base.

Complete step by step answer:
Buffer solution resists a change in pH upon dilution or upon the addition of small amounts of acid or base to them. Buffer index can be defined as the differential ratio of the increase in the amount of strong acid or strong base added with respect to pH variation.
We know the formula:
pOH=pKb+log[Salt][Base]pOH=p{{K}_{b}}+\log \dfrac{[Salt]}{[Base]}
Now suppose 1 ml of 1 M HClHCl which represents 0.001 mol of HClHCl is added which converts NH4OHN{{H}_{4}}OH into NH4ClN{{H}_{4}}Cl.
Now, we can say that NH4ClN{{H}_{4}}Cl= 0.1+0.001=0.101M0.1+0.001=0.101M
NH4OHN{{H}_{4}}OH= 0.10.001=0.099M0.1-0.001=0.099M
By putting the values in the above formula, where the value of pKbp{{K}_{b}}= 4.74
pOH=4.74+log[0.101][0.099]=4.74+0.0086pOH=4.74+\log \dfrac{[0.101]}{[0.099]}=4.74+0.0086
This defines the change in pH is 0.0086 and the formula for buffer index is:
= dndpH\dfrac{dn}{dpH} where dndn= number of moles added i.e. 0.001 and dpHdpH= change in pH = 0.0086
= 0.0010.0086=0.115\dfrac{0.001}{0.0086}=0.115

Hence we can say that buffer index of a buffer of 0.1 M NH4OHN{{H}_{4}}OH and 0.1 M NH4ClN{{H}_{4}}Cl is 0.115, option B is the correct answer.

Note: Buffer capacity can be numerically expressed to be equal with the minimum concentration of strong acid or strong base which cause the changes in buffer’s pH value by one unit. The pH of Buffer Solutions shows minimum change upon the addition of a very small quantity of strong acid or strong base.