Chemistry · 2026
JEE Main · 4 April 2026, Shift 2 · Q55
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…
$\displaystyle 20$ mL of a solution of acetic acid required $\displaystyle 28.4$ mL of $\displaystyle 0.1$ M NaOH for its neutralization. A solution $\displaystyle (\mathrm{X})$ was prepared by mixing $\displaystyle 20$ mL of the above acetic acid and $\displaystyle 14.2$ mL of $\displaystyle 0.1$ M NaOH solution. What is the pH of the solution (X)? $\displaystyle \left(\mathrm{pK}_{\mathrm{a}}\right.$ value of acetic acid is $\displaystyle 4.75 )$.
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 4.75$
More from Ionic Equilibrium
- M_3 A_2 is a sparingly soluble salt of molar mass y g mol^-1 and solubility x g L^-1. The ratio of the molar…2026
- Identify the colour of compound ' X ' in the sequence of the reaction.2026
- Given below are two statements: Statement I: Sodium dichromate and potassium dichromate are classified as…2026
- The pH of a solution obtained by mixing 5 mL of 0.1 M NH_4 OH solution with 250 mL of 0.1 M NH_4 Cl solution…2026
- Given is a concentrated solution of a weak electrolyte A_x B_y of concentration ' c ' and dissociation…2026
- Arrange the following resultant mixtures in increasing order of their pH values A. 10 mL 0.2 M Ca(OH)_2+25 mL…2026
- At 25°C, 20.0 mL of 0.2 M weak monoprotic acid HX is titrated against 0.2 M NaOH. The pH of the solution at…2026
- The first and second ionization constants of a weak dibasic acid H_2 A are 8.1 × 10^-8 and 1.0 × 10^-13…2026
JEE Main 2026 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.