JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
\(20\) mL of a solution of acetic acid required \(28.4\) mL of \(0.1\) M NaOH for its neutralization. A solution (X) was prepared by mixing \(20\) mL of the above acetic acid and \(14.2\) mL of \(0.1\) M NaOH solution. What is the pH of the solution (X)? (\(pK_a\) value of acetic acid is \(4.75\)).
- A \(7.0\)
- B \(4.75\)
- C \(3.5\)
- D \(4.82\)
Answer & Solution
Correct Answer
(B) \(4.75\)
Step-by-step Solution
Detailed explanation
For complete neutralization of \(20\) mL of acetic acid, the millimoles of NaOH required is: \(28.4 \times 0.1 = 2.84\) mmol Thus, \(20\) mL of the acetic acid solution contains \(2.84\) mmol of \(CH_3COOH\). For the preparation of solution (X), the millimoles of NaOH added is:…
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