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Question: Useful buffer range of weak acid HA \[{K_a} = {10^{ - 5}}\] is: A. 5 to 7 B.4 to 6 C.3 to 5 ...

Useful buffer range of weak acid HA Ka=105{K_a} = {10^{ - 5}} is:
A. 5 to 7
B.4 to 6
C.3 to 5
D.None of these

Explanation

Solution

vHint: Ka{K_a}is the acid dissociation constant. It helps to determine the strength of the acid. The pHpHrange where the buffer neutralizes the added acids and bases is called buffer range. While doing so, the buffer tries to maintain the constantpHpH.
Formula used:

pH=pKa+log[conjugateBase][acid] pKa=log10Ka  pH = p{K_a} + \log \dfrac{{[conjugateBase]}}{{[acid]}} \\\ p{K_a} = - {\log _{10}}{K_a} \\\

Complete step by step answer:
Given,Ka=105{K_a} = {10^{ - 5}}
Using the relation, pKa=log10Kap{K_a} = - {\log _{10}}{K_a}
We get pKa=log10Ka=log10105=5p{K_a} = - {\log _{10}}{K_a} = - {\log _{10}}{10^{ - 5}} = 5
Now, we will use this value to determine the potential of hydrogenpHpHby using the formula:
pH=pKa+log[conjugateBase][acid]pH = p{K_a} + \log \dfrac{{[conjugateBase]}}{{[acid]}}
For an acid HA,
Case 1: [conjugateBase][acid]=110\dfrac{{[conjugateBase]}}{{[acid]}} = \dfrac{1}{{10}}
Substituting this value,
pH=pKa+log110pH = p{K_a} + \log \dfrac{1}{{10}}
We get,pH=5+log0.1pH = 5 + \log 0.1
pHpH=5-1=4 …(I)
Case 2: [conjugateBase][acid]=101\dfrac{{[conjugateBase]}}{{[acid]}} = \dfrac{{10}}{1}
Substituting this value
pH=5+log101pH = 5 + \log \dfrac{{10}}{1}
We get,
pHpH=5+1=6 …..(II)
So, Useful buffer range of weak acid is 4 to 6.

Hence, option B. (4 to 6) is the correct option to the given question.

Note:
Lesser the value ofpKap{K_a}, stronger is the acid. Temperature and concentration can affect thepKap{K_a} Buffer usually maintains thepHpH of the system and resists the change in the concentration of H+{H^ + } ions in the system or solution.