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
- A \(1.14\)
- B \(1.78\)
- C \(2.34\)
- D \(3.02\)
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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