An acid-base indicator HIn in a buffer of pH = pK In -1 is ionized to the extent of about

Question

An acid-base indicator HIn in a buffer of pH = pK In -1 is ionized to the extent of about  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

pH = pK textIn +([ textIn -]/[ textHIn ])=( pK text In -1) log ([ textIn -]/[ textHin ])=-1 longrightarrow ([ textIn -]/[ textHln ])=0.1 % ionized =(0.1 × 100/0.1+1)=(10/1.1)=9.1: 9.0

Q. An acid-base indicator HIn in a buffer of $p H = p K_{I n} - 1$ is ionized to the extent of about

$p H = p K_{I n} + \frac{[ I n ^{-} ]}{[ H I n ]} = (p K_{I n} - 1)$
$lo g \frac{[ I n ^{-} ]}{[ H in ]} = - 1 ⟶ \frac{[ I n ^{-} ]}{[ H l n ]} = 0.1$
$%$ ionized $= \frac{0.1 \times 100}{0.1 + 1} = \frac{10}{1.1} = 9.1 : 9.0$

Q. An acid-base indicator HIn in a buffer of pH=pKIn​−1 is ionized to the extent of about

pH=pKIn​+[HIn][In−]​=(pKIn​−1) log[Hin][In−]​=−1⟶[Hln][In−]​=0.1 % ionized =0.1+10.1×100​=1.110​=9.1:9.0

Q. An acid-base indicator HIn in a buffer of $p H = p K_{I n} - 1$ is ionized to the extent of about

$p H = p K_{I n} + \frac{[ I n ^{-} ]}{[ H I n ]} = (p K_{I n} - 1)$
$lo g \frac{[ I n ^{-} ]}{[ H in ]} = - 1 ⟶ \frac{[ I n ^{-} ]}{[ H l n ]} = 0.1$
$%$ ionized $= \frac{0.1 \times 100}{0.1 + 1} = \frac{10}{1.1} = 9.1 : 9.0$