A certain acidic indicator has pH =5.48. 75 % of the indicator exists in its ionised from. Calculate Ka of the indicator,

Question

A certain acidic indicator has pH =5.48. 75 % of the indicator exists in its ionised from. Calculate Ka of the indicator, (A) 1 × 10-5 (B) 2 × 10-5 (C) 1 × 10-3 (D) 4 × 10-5

Tardigrade Admin · Accepted Answer

pH = pk In + log (( In -/ H In )) 5.48= pk In + log ((75/25)) pk In =5.48- log 3=5.48-0.48=5 k In =1 × 10-5

Q. A certain acidic indicator has $p H = 5.48. 75%$ of the indicator exists in its ionised from. Calculate $K a$ of the indicator,

$1 \times 1 0^{- 5}$

$2 \times 1 0^{- 5}$

$1 \times 1 0^{- 3}$

$4 \times 1 0^{- 5}$

$p H = p k_{I n} + lo g (\frac{I n ^{-}}{H I n})$
$5.48 = p k_{I n} + lo g (\frac{75}{25})$
$p k_{I n} = 5.48 - lo g 3 = 5.48 - 0.48 = 5$
$k_{I n} = 1 \times 1 0^{- 5}$

Q. A certain acidic indicator has pH=5.48.75% of the indicator exists in its ionised from. Calculate Ka of the indicator,

1×10−5

2×10−5

1×10−3

4×10−5