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

\(0.01 \mathrm{M} \mathrm{HCl}\) के \(600 \mathrm{~mL}\) को \(400 \mathrm{~mL} 0.01 \mathrm{M}\) \(\mathrm{H}_2 \mathrm{SO}_4\) के साथ मिश्रित करने पर प्राप्त मिश्रण की \(\mathrm{pH}\) है। _________ \(\times 10^{-2}\). (निकटतम पूर्णाक में)) [दिया है \(\log 2=0.30, \quad \log 3=0.48\) \(\log 5=0.69 \quad \log 7=0.84\) \(\log 11=1.04]\)

  1. A \(185\)
  2. B \(186\)
  3. C \(184\)
  4. D \(183\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(186\)

Step-by-step Solution

Detailed explanation

Total milimoles of \(H ^{+}=(600 \times 0.01)+(400 \times 0.01 \times 2)\) \(=14\) \(\left[ H ^{+}\right]=\frac{14}{1000}=14 \times 10^{-3}\) \(pH =3-\log 14\) \(=1.86\) \(=186 \times 10^{-2}\)
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