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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Ionic Equilibrium

NTA Abhyas 2020 · Chemistry · Ionic Equilibrium

For the indicator, HIn; the ratio& [ In ^ - ] [HIn] is 7 .0 at pH...

For the indicator, HIn; the ratio& [In−][HIn]\dfrac{\left[{In}^{-}\right]}{\left[\text{HIn}\right]} is 7.07 .0 at pH of 4.34 .3. KIn{K}_{\text{In}} for the indicator is

[ Given: log⁡7=0.845\log 7 = 0 .845 and Antilog (−3.455)=3.5×10−4( - 3.455 ) = 3.5 \times {10}^{- 4} ]

Options

  1. A3.5×10−43 \text{.} 5 \times 1 {0}^{- 4}
  2. B3.5×10−53 \text{.} 5 × 1 {0}^{- 5}
  3. C3.5×10−23 \text{.} 5 × 1 {0}^{- 2}
  4. D3.5×10−33 \text{.} 5 × 1 {0}^{- 3}

Correct answer

A. 3.5×10−43 \text{.} 5 \times 1 {0}^{- 4}

Step-by-step solution

For weak organic acid indicators

\text{pH} = {\text{pK}_{\text{In}} + \text\log\dfrac{\left[{\text{In}^{−}\right]}{\left[\text{HIn}\right]}

\text{4}\text{.3} = {\text{pK}_{\text{In}} + \text{\log 7}

{\text{pK}_{\text{In}} = 4.3 - 0.845 = 3.455

{\text{pK}_{\text{In}} = - {\text\log}_{\text{10}} {K}_{\text{In}}

{K}_{\text{In}} = \text{Antilog} \left(- {\text{pK}_{\text{In}}\right)

KIn{K}_{\text{In}}=Antilog(−3.455)= \text{Antilog} \left(-3.455\right)

=3.5×10−4= 3.5 \times {10}^{- 4}
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