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JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)

Two solutions \(A\) and \(B\), each of \(100\; L\) was made by dissolving \(4 \;\mathrm{g}\) of \(\mathrm{NaOH}\) and \(9.8 \;\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in water, respectively. The \(\mathrm{pH}\) of the resultant solution obtained from mixing \(40\; \mathrm{L}\) of solution \(A\) and \(10\; \mathrm{L}\) of solution \(\mathrm{B}\) is

  1. A \(7.3\)
  2. B \(8.7\)
  3. C \(9.6\)
  4. D \(10.6\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(10.6\)

Step-by-step Solution

Detailed explanation

\(4 \mathrm{gm}\) of \(\mathrm{NaOH}\) in \(100 \mathrm{L}\) sol. \(\Rightarrow 10^{-3} \mathrm{M}\) sol. \(9.8 \mathrm{gm}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in \(100 \mathrm{L}\) sol. \(\Rightarrow 10^{-3} \mathrm{M}\) sol. Mixture : \(40 L\) of…
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