Calculate pH at which an acid indicator HIn with concentration 0.1 M changes its colour ( K a . for HIn =.1 × 10-5)

Question

Calculate pH at which an acid indicator HIn with concentration 0.1 M changes its colour ( K a . for HIn =.1 × 10-5) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Indicator equation is: pH = pK textIn + log ([ textIn -]/[ textH In ]) Colour change takes place when [ textIn -]=[ textH In ] i.e., when pH = pK textIn =5

Q. Calculate $p H$ at which an acid indicator HIn with concentration $0.1 M$ changes its colour $(K_{a}$ for HIn $= 1 \times 1 0^{- 5})$

Indicator equation is: $p H = p K_{I n} + lo g \frac{[ I n ^{-} ]}{[ H I n ]}$
Colour change takes place when $[I n^{-}] = [H I n]$
i.e., when $p H = p K_{I n} = 5$

Q. Calculate pH at which an acid indicator HIn with concentration 0.1M changes its colour (Ka​ for HIn =1×10−5)

Indicator equation is: pH=pKIn​+log[HIn][In−]​ Colour change takes place when [In−]=[HIn] i.e., when pH=pKIn​=5

Q. Calculate $p H$ at which an acid indicator HIn with concentration $0.1 M$ changes its colour $(K_{a}$ for HIn $= 1 \times 1 0^{- 5})$

Indicator equation is: $p H = p K_{I n} + lo g \frac{[ I n ^{-} ]}{[ H I n ]}$
Colour change takes place when $[I n^{-}] = [H I n]$
i.e., when $p H = p K_{I n} = 5$