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

\(200\; mL\) of \(0.01\; M\; HCl\) is mixed with \(400 \;mL\) of \(0.01\; M\; H _{2} SO _{4}\). The \(pH\) of the mixture is

  1. A \(1.14\)
  2. B \(1.78\)
  3. C \(2.34\)
  4. D \(3.02\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(1.78\)

Step-by-step Solution

Detailed explanation

\(HCl + H _{2} SO _{4}\) \(\left[ H ^{+}\right]=\frac{(0.01 \times 200)+(0.01 \times 2 \times 400)}{600}\) \(=\frac{2+8}{600}=\frac{10}{600}=\frac{1}{60}\) \(pH =-\log [\frac{1}{60}]\) \(=1.78\)
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