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Question: The pH range of a basic indicator ( \(\ln OH\) ) is \(3.4\) to \(4.6\). For what minimum ratio of \(...

The pH range of a basic indicator ( ln⁡OH\ln OH ) is 3.43.4 to 4.64.6. For what minimum ratio of [In+]InOH\dfrac{{[I{n^ + }]}}{{InOH}} does the solution appear in the colour of In?
A) 4
B) 2
C) 8
D) 16

Explanation

Solution

pH is the value of negative log of [H+][{H^ + }] i.e. pH is given as −log⁡[H+] - \log [{H^ + }] and pOH is the value of negative log of [OH−][O{H^ - }] and pOH is given as −log⁡[OH−] - \log [O{H^ - }]. The relationship between pH and pOH is given as: pH=14−pOHpH = 14 - pOH

Complete answer:
In chemistry indicators are used to indicate the presence of any compound in a mixture or solution by means of change in physical appearances like change in colour. Phenolphthalein, methyl blue, methyl orange, cresol red are some commonly used indicators.
The colour of the indicator changes when it reaches its maximum pH range. In the given question the range given to us is 3.43.4 to 4.64.6 . After the pH range of 4.64.6 is crossed the indicator shows the colour change.
Initially the indicator when mixed in a solution exists as a complex/compound InOHInOH . This is a weak compound and gets dissociated easily. The dissociation can be given as:
InOH⇌In++OH−InOH \rightleftharpoons I{n^ + } + O{H^ - }
From the above equation the relationship between pOH and pKbp{K_b} can be given as:
pOH=pKb+log⁡(In+InOH)pOH = p{K_b} + \log \left( {\dfrac{{I{n^ + }}}{{InOH}}} \right)
First, we need to find the value of pKInp{K_{In}} . pKInp{K_{In}} can be found out as the mean of pOH range of the indicator. Given to us is the pH range of the indicator. Converting it into the pOH range we get,
pOH range =9.4−10.6 = 9.4 - 10.6 . Hence after pOH of 10.610.6 is crossed the indicator will show the colour change. The pKInp{K_{In}} can be given as:
pKb=10.6+9.42=10p{K_b} = \dfrac{{10.6 + 9.4}}{2} = 10
Substituting the value of pKbp{K_b} in the above equation to find the ratio of [In+]InOH\dfrac{{[I{n^ + }]}}{{InOH}}
10.6=10+log⁡(In+InOH)10.6 = 10 + \log \left( {\dfrac{{I{n^ + }}}{{InOH}}} \right)
0.6=log⁡(In+InOH)0.6 = \log \left( {\dfrac{{I{n^ + }}}{{InOH}}} \right)
(In+InOH)=AL(0.6)\left( {\dfrac{{I{n^ + }}}{{InOH}}} \right) = AL(0.6)
[In]+InOH=4\dfrac{{{{[In]}^ + }}}{{InOH}} = 4
Hence the correct answer is Option (A).

Note:
Every indicator has its own pH range, for example Phenolphthalein has a pH range of 8−108 - 10. We can also solve this question by using the pH values itself. The formula would be:
pH=pKIn+log⁡(In+InOH)pH = p{K_{In}} + \log \left( {\dfrac{{I{n^ + }}}{{InOH}}} \right)
pKIn=3.4+4.62p{K_{In}} = \dfrac{{3.4 + 4.6}}{2}