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

NTA Abhyas 2020 · Chemistry · Ionic Equilibrium

What is [H^ + ] in a solution that is 0.01M in HCN and 0.02M in...

What is [H+]\left[H^{+}\right] in a solution that is 0.01MinHCNand0.02MinNaCN?0.01M in HCN and 0.02M in \text{NaCN?}
br /> (Ka( {K}_{a}for HCN=6.2×10−10)HCN = 6.2 \times 1{0}^{- 10} )

Options

  1. A3.1×10103.1 \times 1{0}^{10}
  2. B6.2×1056.2 \times 1{0}^{5}
  3. C6.2×10−106.2 \times 1{0}^{- 10}
  4. D3.1×10−103.1 \times 1{0}^{- 10}

Correct answer

D. 3.1×10−103.1 \times 1{0}^{- 10}

Step-by-step solution

Ka=[H+][CN−][HCN]{K}_{a} \text{=}\dfrac{\left[H^{+} \left]\right[{\text{CN}}^{−}\right]}{\left[\text{HCN}\right]}
6.2×10−10=[H+][0.02][0.01]6.2 \times 1{0}^{- 10} = \dfrac{[ H^{+} ] [ 0.02 ]}{[ 0.01 ]}
\left[\text{NaCN}\right]\text{=} \left[{\text{CN}}^{−}\right]duetocommonioneffect<br/>[H+]=6.2×10−10×0.010.02=due to common ion effect<br />\left[H^{+}\right] =\dfrac{6.2 \times 1{0}^{- 10} \times 0.01}{0.02} =3.1 \times10−101{0}^{- 10}
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