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 K In -1 - 10-5A- 1B- 0-1C- 2D- 0-2

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

Potassium Permanganate

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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Potassium Permanganate

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