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NEET CHEMISTRYEquilibriumMedium

Question

The pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed is:

A

12.65

B

2.04

C

7.01

D

1.35

Step-by-Step Solution

Let the volume of each solution be VV. Moles of NaOHNaOH = 0.1×V=0.1V0.1 \times V = 0.1V Moles of HClHCl = 0.01×V=0.01V0.01 \times V = 0.01V The neutralization reaction is NaOH+HClNaCl+H2ONaOH + HCl \rightarrow NaCl + H_2O. Since NaOHNaOH and HClHCl react in a 1:1 molar ratio, HClHCl is the limiting reagent. Moles of NaOHNaOH left unreacted = 0.1V0.01V=0.09V0.1V - 0.01V = 0.09V Total volume of the resulting mixture = V+V=2VV + V = 2V Concentration of OHOH^- in the final solution = 0.09V2V=0.045 M\frac{0.09V}{2V} = 0.045\text{ M} pOH=log[OH]=log(0.045)=log(4.5×102)=2log(4.5)=20.653=1.347pOH = -\log[OH^-] = -\log(0.045) = -\log(4.5 \times 10^{-2}) = 2 - \log(4.5) = 2 - 0.653 = 1.347 pH=14pOH=141.347=12.65312.65pH = 14 - pOH = 14 - 1.347 = 12.653 \approx 12.65.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Equilibrium. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYEquilibriumresultingsolutionvolumes

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The following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations. pH of which one of them will be equal to 1?

A.$60 \text{ mL } \frac{M}{10} \text{ HCl } + 40 \text{ mL } \frac{M}{10} \text{ NaOH}$
B.$55 \text{ mL } \frac{M}{10} \text{ HCl } + 45 \text{ mL } \frac{M}{10} \text{ NaOH}$
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The tendency of $BF_3$, $BCl_3$ and $BBr_3$ to behave as Lewis acid decreases in the sequence:

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A.$2 \times 10^{-4} \text{ mol/L}$
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