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

NTA Abhyas 2020 · Chemistry · Ionic Equilibrium

20ml of 0.1MNaOH is added to 30ml0.1M NaOH30ml0.2 M CH_ 3 COOH (...

20ml20ml of 0.1MNaOH0.1MNaOH is added to 30\text{ml}0.1\text{M NaOH}30\text{ml}0.2{\text{M CH}_{3} \text{COOH}\left({\text{pK}_{a} =4.74\right),calculatethe, calculate the\text{pH}oftheresultingsolution?[given:of the resulting solution? [given:\log 2 = 0.3 ]

Options

  1. A3.44
  2. B4.01
  3. C4.44
  4. D4.71

Correct answer

C. 4.44

Step-by-step solution

{NaOH+CH3COOH→CH3COONa+H2Ommol26LR−42\begin{cases} & NaOH & + & C{H}_{3}COOH & \to & C{H}_{3}COONa & + & {H}_{2}O \\ mmol & 2 & & 6 & & & & \\ & LR & & & & & & \\ & - & & 4 & & 2 & & \end{cases}

The resulting solution is an acidic buffer. so pH=pKa+log[Salt][Acid]\text{pH} = {\text{pK}}_{a} + log\dfrac{\left[\text{Salt}\right]}{\left[\text{Acid}\right]}

=4.71+log24= 4.71 + log \dfrac{2}{4}

=4.44= 4.44
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