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Question: An acid base indicator has \(Ka = 3 \times {10^{ - 5}}\),the acid form of indicator is red and the b...

An acid base indicator has Ka=3×10−5Ka = 3 \times {10^{ - 5}},the acid form of indicator is red and the basic form of indicator is blue. By how much must the pHpHchange in order to change the indicator from 75% red to 75% blue?
A. 7.91×10−57.91 \times {10^{ - 5}}
B. 0.950.95
C. 5×10235 \times {10^{23}}
D. 0.90.9

Explanation

Solution

To solve the given question 1st we need to find out the pHpH at different concentrations, after finding out pHpH we will subtract it and will get the final solution. The relationship between pHpH and Ka{K_a} is given by:
pH=pKa+log⁡[ln⁡−][Hln⁡]pH = p{K_a} + \log \dfrac{{[{{\ln }^ - }]}}{{[{\text{H}}\ln ]}}
Here pKap{K_a} represents the acidic strength, if lower is the value stronger is the acid and vice-versa..

Complete step by step answer:
According to the given question
The Final concentration of [Hln⁡][{\text{H}}\ln ] =75%
The Initial concentration of [ln⁡−][{\ln ^ - }] =25%
The Final concentration of [Hln⁡][{\text{H}}\ln ] =25%
The Initial concentration of [ln⁡−][{\ln ^ - }] =75%
Therefore now by substituting the above formula in the above relationship, let us find out the value of pHpH at two different concentrations.
pH1=−log⁡(3×105)+log⁡75/25p{H_1} = - \log (3 \times {10^5}) + \log 75/25
⇒pH1= 5\Rightarrow p{H_1} = {\text{ }}5
⇒[H1+]=10−5M\Rightarrow [H_1^ + ] = {10^{ - 5}}M … (i)
pH2=−log⁡(3×105)+log⁡25/75p{H_2} = - \log (3 \times {10^5}) + \log 25/75
⇒pH2= - 4.045\Rightarrow p{H_2} = {\text{ - 4}}{\text{.045}}
⇒[H2+]=8.91×10−5M\Rightarrow [H_2^ + ] = 8.91 \times {10^{ - 5}}M … (ii)
Now subtracting equation (ii) and (i), we get
[H2+]−[H1+]=7.91×10−5M[H_2^ + ] - [H_1^ + ] = 7.91 \times {10^{ - 5}}M

So, the correct answer is Option A .

Additional Information:
pKpK value: pp stands for negative logarithm. Just as H+{H^ + } and OH−O{H^ - } ion concentration range over many negative power of 10, it is convenient to express them as pHpHor pOHpOH, the dissociation constant (K)(K) value also range over many negative power of 10 and it is convenient to write them as pKpK. Thus, pKpK is the negative logarithm of dissociation constant. Sometimes pHpH of acid comes more than 7 and that of base comes less than 7. It shows that the solution is very dilute.

Note:

Weak acids have higher pKap{K_a} values. Similarly, weak bases have higher pKbp{K_b} values.
A solution of pHpH = 1 has a hydrogen ion concentration 100 times that of a solution of pHpH= 3.
pHpH of zero is obtained in 1N solution of strong acid. In case the concentration is 2N, 3N, 4N, etc. the respective pHpH value will be negative.