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

NTA Abhyas 2020 · Chemistry · Ionic Equilibrium

When 10mL of 0.1M acetic acid (pK_ a =5) is titrated against 10mL...

When 10mL\text{10}\text{mL} of 0.1M\text{0}\text{.1}\text{M} acetic acid {\text{(pK}_{a} \text{=}\text{5)} is titrated against 10mL\text{10}\text{mL} of 0.1M\text{0}\text{.1}\text{M} ammonia solution {\text{(pK}_{b} = 5 ) , the equivalent point will occur at pH\text{pH}:

Options

  1. A5
  2. B6
  3. C7
  4. D9

Correct answer

C. 7

Step-by-step solution

\begin{cases}{\text{CH}_{3} \text{COOH}\text{(aq)} & \text{+} & {\text{NH}_{3} \text{(aq)} & \to & {\text{CH}_{3} {\text{COONH}_{4} \text{(aq)} & & \\ \text{1}\text{m}\text{mol} & & \text{1}\text{m}\text{mol} & & \text{1}\text{m}\text{mol} & & \\ & & & & & & \end{cases}<br />\left[{\text{CH}_{3} {\text{COONH}_{4}\right]\text{=}\dfrac{\text{1}\text{mM}}{\text{20}\text{mL}} \text{=}\text{0}\text{.05 M}
Salt of weak Acid + weak base.
pH=12(pKw+pKa−pKb)<br/>=12(14+5−5)=7\text{pH} = \dfrac{1}{2} \left(\text{p} {K}_{w} + \text{p} {K}_{a} - \text{p} {K}_{b}\right)<br />=\dfrac{1}{2} \left(1 4 + 5 - 5\right) = 7
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