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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 the [ OH ^ − ] concentration of a 0.04M solution of CH_ 3...

What is the [OH−]\left[{\text{OH}}^{−}\right] concentration of a 0.04M\text{0}\text{.04}\text{M} solution of {\text{CH}_{3} \text{COONa?}
{\text{[K}}_{a} \text{of}{\text{CH}_{3} \text{COONa}\text{=}\text{2}\text{\times}{\text{10}}^{- \text{5}} \text{,}\text{\log 2}\text{=}\text{0}\text{.30,}\text{\log 5}\text{=}\text{0}\text{.69,}{\text{10}}^{\text{0}\text{.7}} ≅ \text{5]}

Options

  1. A5×10−65 \times {10}^{- 6}
  2. B6×10−66 \times {10}^{- 6}
  3. C2×10−92 \times {10}^{- 9}
  4. D3×10−93 \times {10}^{- 9}

Correct answer

A. 5×10−65 \times {10}^{- 6}

Step-by-step solution

{\text{pK}_{a} \text{=} - \text\log{K}_{a} \text{=} - \text\log\text{(2} \times {\text{10}^{- 5} )
=5−log⁡2=5−0.3=4.7= 5 - \log 2 = 5 - 0.3 = 4.7
\text{pH}\text{=}\dfrac{1}{2} \left({\text{pK}_{w} \text{+}{\text{pK}_{a} \text{+}\text\log\text{C}\right)<br /> pH = 12[14+4.7+log⁡(0.04)]\dfrac{1}{2} [14 + 4.7 + \log \left(0.04\right) ]
pH=8.63pH = 8.63
pOH=14−8.63≅5.3<br/>[OH−]=5×10−6\text{pOH}\text{=}\text{14} - \text{8}\text{.63} ≅ \text{5}\text{.3}<br />\left[{OH}^{-}\right] =5 \times {10}^{- 6}
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