Consider an indicator HIn- Find the change in pH i-e- - pH of an acidic indicator when the ratio -In-HIn- changes from 0-2 to 2-Given KIn-1- 10-5

K_in=1× 10^ 5 pK_in= log(K_in) pK_in= log(1× 10^ 5) pK_in=5 pH=pK_in+log([In^ ][HIn]) Case 1 : pH_1=5+log(0.2)=5 0.7=4.3 pH_1=4.3 Case 2 : pH_2=5+log(2)=5+0.3 ...

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Standard XII

Chemistry

pH the Power of H

Consider an i...

Question

Consider an indicator $H I n$ . Find the change in $p H$ i.e. $Δ p H$ of an acidic indicator when the ratio $\frac{[I n^{-}]}{[H I n]}$ changes from $0.2$ to $2$ .
Given $K_{I n} = 1 \times 10^{- 5}$

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Solution

The correct option is B $1$
$K_{i n} = 1 \times 10^{- 5} p K_{i n} = - l o g (K_{i n}) p K_{i n} = - l o g (1 \times 10^{- 5}) p K_{i n} = 5$
$p H = p K_{i n} + log (\frac{[I n^{-}]}{[H I n]})$
Case 1 :
$p H_{1} = 5 + l o g (0.2) = 5 - 0.7 = 4.3 p H_{1} = 4.3$

Case 2 :
$p H_{2} = 5 + l o g (2) = 5 + 0.3 = 5.3 p H_{2} = 5.3$
Change in $p H :$
$Δ p H = p H_{2} - p H_{1} Δ p H = 5.3 - 4.3 = 1$

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Similar questions

Q. Consider an indicator

H I n

. Find the change in

p H

i.e.

Δ p H

of an acidic indicator when the ratio

\frac{[I n^{-}]}{[H I n]}

changes from

0.2

2

.
Given

K_{I n} = 1 \times 10^{- 5}

Q. Calculate the

p H

at which an acid indicator

(H I n)

having

K_{I n} = 10^{- 5}

changes its colour.
Given, the

H I n

concentration is

10^{- 3} M

Q. The rapid change in

p H

near the equivalence point of an acid-base titration is the basis of indicator detection.

p H

of the solution is related to ratio of the concentrations of the acidic form (

H I n

) and basic form (

I n^{-})

of the indicator by the expression :

Q. The indicator constant for an acidic indicator , HIn is

5 \times 10^{- 6} M

. this indicator appears only in the colour of acidic from when

\frac{[I n^{-}]}{[H I n]} \leq \frac{1}{20}

and it appears only in the colour of basic form when

\frac{[H I n]}{[I n^{-}]} \leq \frac{1}{40} .

the pH range of indicator is:

Q. An acid type indicator,

H I n

, differs in colour from its conjugate base

{I n}^{-}

. The human eye is sensitive to colour differences only when the ratio

[{I n}^{-}] / [H I n]

is greater than

10

or smaller than

0.1

. What should be the minimum change in pH of solution to observe a complete colour change?

(K_{a} = 10^{- 5})

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pH the Power of H

Standard XII Chemistry

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Consider an indicator HIn. Find the change in pH i.e. ΔpH of an acidic indicator when the ratio [In−][HIn] changes from 0.2 to 2. Given KIn=1×10−5

The correct option is B 1Kin=1×10−5pKin=−log(Kin)pKin=−log(1×10−5)pKin=5 pH=pKin+log([In−][HIn]) Case 1 : pH1=5+log(0.2)=5−0.7=4.3pH1=4.3 Case 2 : pH2=5+log(2)=5+0.3=5.3pH2=5.3 Change in pH: ΔpH=pH2−pH1ΔpH=5.3−4.3=1

Consider an indicator $H I n$ . Find the change in $p H$ i.e. $Δ p H$ of an acidic indicator when the ratio $\frac{[I n^{-}]}{[H I n]}$ changes from $0.2$ to $2$ .
Given $K_{I n} = 1 \times 10^{- 5}$